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Inductor and Capacitor in Parallel

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Inductor and Capacitor in Parallel



The Next CEO of Stack OverflowCapacitor related queryPhysical question on an RLC circuitCapacitor / inductor resonanceWhy do Capacitor Inductor circuits Oscillate instead of reaching equilibrium?Current in inductor just after switch is closedWhy do switch burns out due to charged Inductor and capacitor while performing switching operation?Charging current of capacitorRC Discharge/ChargingWhy does discharging a capacitor give a higher power output than using a battery?Inductor and Capacitor with a DC supply










2












$begingroup$


I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!



enter image description here



So, let's say we close the switch. From what I understand, the capacitor will charge up to $10V$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $0.2A$ flowing through the resistor.



But what happens if we leave the switch closed for a long time?



My book says the current through the inductor would pick up to $0.2A$, while the current through the capacitor drops to $0A$.



However, that seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?



In terms of voltage drop - if the current through the resistor were ever to reach $0.2A$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.



Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?



Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor? Thanks!










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Shouldn't I = 0.2A in the beginning
    $endgroup$
    – Starboy
    13 hours ago










  • $begingroup$
    @Starboy fixing it now! Thanks!
    $endgroup$
    – Joshua Ronis
    11 hours ago















2












$begingroup$


I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!



enter image description here



So, let's say we close the switch. From what I understand, the capacitor will charge up to $10V$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $0.2A$ flowing through the resistor.



But what happens if we leave the switch closed for a long time?



My book says the current through the inductor would pick up to $0.2A$, while the current through the capacitor drops to $0A$.



However, that seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?



In terms of voltage drop - if the current through the resistor were ever to reach $0.2A$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.



Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?



Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor? Thanks!










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Shouldn't I = 0.2A in the beginning
    $endgroup$
    – Starboy
    13 hours ago










  • $begingroup$
    @Starboy fixing it now! Thanks!
    $endgroup$
    – Joshua Ronis
    11 hours ago













2












2








2


2



$begingroup$


I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!



enter image description here



So, let's say we close the switch. From what I understand, the capacitor will charge up to $10V$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $0.2A$ flowing through the resistor.



But what happens if we leave the switch closed for a long time?



My book says the current through the inductor would pick up to $0.2A$, while the current through the capacitor drops to $0A$.



However, that seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?



In terms of voltage drop - if the current through the resistor were ever to reach $0.2A$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.



Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?



Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor? Thanks!










share|cite|improve this question











$endgroup$




I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!



enter image description here



So, let's say we close the switch. From what I understand, the capacitor will charge up to $10V$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $0.2A$ flowing through the resistor.



But what happens if we leave the switch closed for a long time?



My book says the current through the inductor would pick up to $0.2A$, while the current through the capacitor drops to $0A$.



However, that seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?



In terms of voltage drop - if the current through the resistor were ever to reach $0.2A$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.



Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?



Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor? Thanks!







homework-and-exercises electricity electric-circuits capacitance electromagnetic-induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 11 hours ago







Joshua Ronis

















asked 13 hours ago









Joshua RonisJoshua Ronis

1,3322520




1,3322520







  • 1




    $begingroup$
    Shouldn't I = 0.2A in the beginning
    $endgroup$
    – Starboy
    13 hours ago










  • $begingroup$
    @Starboy fixing it now! Thanks!
    $endgroup$
    – Joshua Ronis
    11 hours ago












  • 1




    $begingroup$
    Shouldn't I = 0.2A in the beginning
    $endgroup$
    – Starboy
    13 hours ago










  • $begingroup$
    @Starboy fixing it now! Thanks!
    $endgroup$
    – Joshua Ronis
    11 hours ago







1




1




$begingroup$
Shouldn't I = 0.2A in the beginning
$endgroup$
– Starboy
13 hours ago




$begingroup$
Shouldn't I = 0.2A in the beginning
$endgroup$
– Starboy
13 hours ago












$begingroup$
@Starboy fixing it now! Thanks!
$endgroup$
– Joshua Ronis
11 hours ago




$begingroup$
@Starboy fixing it now! Thanks!
$endgroup$
– Joshua Ronis
11 hours ago










5 Answers
5






active

oldest

votes


















3












$begingroup$


My book says the current through the inductor would pick up to 0.2A,
while the current through the capacitor drops to 0A.




This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.



Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.



Now, recall that the voltage across an (ideal) inductor is given by



$$v_L = Lfracdi_Ldt$$



and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.



For the (ideal) capacitor, the current through is given by



$$i_C = C fracdv_Cdt$$



and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.



In this case, it follows that both the capacitor current and voltage are zero in DC steady state.




Is my book wrong in saying after an infinite amount of time with the
switch closed there will be a 10-volt difference across the capacitor?




Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.




Tangential addendum:




From what I understand, the capacitor will charge up to 10V almost
instantaneously, while no current will flow through the inductor.




That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.



enter image description here






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
      $endgroup$
      – Joshua Ronis
      11 hours ago










    • $begingroup$
      @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
      $endgroup$
      – Gec
      11 hours ago



















    1












    $begingroup$

    From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.






      share|cite|improve this answer









      $endgroup$




















        1












        $begingroup$

        The basic rules for ideal capacitors and inductors are as follows:



        You can’t change the voltage across an ideal capacitor instantaneously.
        You can’t change the current through an ideal inductor instantaneously.



        So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.



        Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor is



        $$V_C=fracQC$$



        Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.



        The capacitor looks like an open circuit.
        The inductor looks like a short circuit.



        That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.



        Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.



        In short, your book is correct.



        Hope this helps.






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
          $endgroup$
          – Starboy
          2 hours ago











        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$


        My book says the current through the inductor would pick up to 0.2A,
        while the current through the capacitor drops to 0A.




        This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.



        Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.



        Now, recall that the voltage across an (ideal) inductor is given by



        $$v_L = Lfracdi_Ldt$$



        and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.



        For the (ideal) capacitor, the current through is given by



        $$i_C = C fracdv_Cdt$$



        and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.



        In this case, it follows that both the capacitor current and voltage are zero in DC steady state.




        Is my book wrong in saying after an infinite amount of time with the
        switch closed there will be a 10-volt difference across the capacitor?




        Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.




        Tangential addendum:




        From what I understand, the capacitor will charge up to 10V almost
        instantaneously, while no current will flow through the inductor.




        That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.



        enter image description here






        share|cite|improve this answer











        $endgroup$

















          3












          $begingroup$


          My book says the current through the inductor would pick up to 0.2A,
          while the current through the capacitor drops to 0A.




          This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.



          Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.



          Now, recall that the voltage across an (ideal) inductor is given by



          $$v_L = Lfracdi_Ldt$$



          and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.



          For the (ideal) capacitor, the current through is given by



          $$i_C = C fracdv_Cdt$$



          and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.



          In this case, it follows that both the capacitor current and voltage are zero in DC steady state.




          Is my book wrong in saying after an infinite amount of time with the
          switch closed there will be a 10-volt difference across the capacitor?




          Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.




          Tangential addendum:




          From what I understand, the capacitor will charge up to 10V almost
          instantaneously, while no current will flow through the inductor.




          That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.



          enter image description here






          share|cite|improve this answer











          $endgroup$















            3












            3








            3





            $begingroup$


            My book says the current through the inductor would pick up to 0.2A,
            while the current through the capacitor drops to 0A.




            This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.



            Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.



            Now, recall that the voltage across an (ideal) inductor is given by



            $$v_L = Lfracdi_Ldt$$



            and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.



            For the (ideal) capacitor, the current through is given by



            $$i_C = C fracdv_Cdt$$



            and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.



            In this case, it follows that both the capacitor current and voltage are zero in DC steady state.




            Is my book wrong in saying after an infinite amount of time with the
            switch closed there will be a 10-volt difference across the capacitor?




            Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.




            Tangential addendum:




            From what I understand, the capacitor will charge up to 10V almost
            instantaneously, while no current will flow through the inductor.




            That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.



            enter image description here






            share|cite|improve this answer











            $endgroup$




            My book says the current through the inductor would pick up to 0.2A,
            while the current through the capacitor drops to 0A.




            This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.



            Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.



            Now, recall that the voltage across an (ideal) inductor is given by



            $$v_L = Lfracdi_Ldt$$



            and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.



            For the (ideal) capacitor, the current through is given by



            $$i_C = C fracdv_Cdt$$



            and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.



            In this case, it follows that both the capacitor current and voltage are zero in DC steady state.




            Is my book wrong in saying after an infinite amount of time with the
            switch closed there will be a 10-volt difference across the capacitor?




            Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.




            Tangential addendum:




            From what I understand, the capacitor will charge up to 10V almost
            instantaneously, while no current will flow through the inductor.




            That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.



            enter image description here







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            Alfred CentauriAlfred Centauri

            48.6k350153




            48.6k350153





















                2












                $begingroup$

                I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                  $endgroup$
                  – Joshua Ronis
                  11 hours ago










                • $begingroup$
                  @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                  $endgroup$
                  – Gec
                  11 hours ago
















                2












                $begingroup$

                I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                  $endgroup$
                  – Joshua Ronis
                  11 hours ago










                • $begingroup$
                  @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                  $endgroup$
                  – Gec
                  11 hours ago














                2












                2








                2





                $begingroup$

                I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?






                share|cite|improve this answer









                $endgroup$



                I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 11 hours ago









                GecGec

                932211




                932211











                • $begingroup$
                  The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                  $endgroup$
                  – Joshua Ronis
                  11 hours ago










                • $begingroup$
                  @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                  $endgroup$
                  – Gec
                  11 hours ago

















                • $begingroup$
                  The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                  $endgroup$
                  – Joshua Ronis
                  11 hours ago










                • $begingroup$
                  @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                  $endgroup$
                  – Gec
                  11 hours ago
















                $begingroup$
                The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                $endgroup$
                – Joshua Ronis
                11 hours ago




                $begingroup$
                The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor?
                $endgroup$
                – Joshua Ronis
                11 hours ago












                $begingroup$
                @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                $endgroup$
                – Gec
                11 hours ago





                $begingroup$
                @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time.
                $endgroup$
                – Gec
                11 hours ago












                1












                $begingroup$

                From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A






                    share|cite|improve this answer









                    $endgroup$



                    From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 11 hours ago









                    StarboyStarboy

                    1568




                    1568





















                        1












                        $begingroup$

                        Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.






                            share|cite|improve this answer









                            $endgroup$



                            Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 11 hours ago









                            PhysicsDavePhysicsDave

                            1,07647




                            1,07647





















                                1












                                $begingroup$

                                The basic rules for ideal capacitors and inductors are as follows:



                                You can’t change the voltage across an ideal capacitor instantaneously.
                                You can’t change the current through an ideal inductor instantaneously.



                                So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.



                                Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor is



                                $$V_C=fracQC$$



                                Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.



                                The capacitor looks like an open circuit.
                                The inductor looks like a short circuit.



                                That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.



                                Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.



                                In short, your book is correct.



                                Hope this helps.






                                share|cite|improve this answer









                                $endgroup$












                                • $begingroup$
                                  Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                  $endgroup$
                                  – Starboy
                                  2 hours ago















                                1












                                $begingroup$

                                The basic rules for ideal capacitors and inductors are as follows:



                                You can’t change the voltage across an ideal capacitor instantaneously.
                                You can’t change the current through an ideal inductor instantaneously.



                                So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.



                                Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor is



                                $$V_C=fracQC$$



                                Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.



                                The capacitor looks like an open circuit.
                                The inductor looks like a short circuit.



                                That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.



                                Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.



                                In short, your book is correct.



                                Hope this helps.






                                share|cite|improve this answer









                                $endgroup$












                                • $begingroup$
                                  Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                  $endgroup$
                                  – Starboy
                                  2 hours ago













                                1












                                1








                                1





                                $begingroup$

                                The basic rules for ideal capacitors and inductors are as follows:



                                You can’t change the voltage across an ideal capacitor instantaneously.
                                You can’t change the current through an ideal inductor instantaneously.



                                So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.



                                Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor is



                                $$V_C=fracQC$$



                                Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.



                                The capacitor looks like an open circuit.
                                The inductor looks like a short circuit.



                                That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.



                                Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.



                                In short, your book is correct.



                                Hope this helps.






                                share|cite|improve this answer









                                $endgroup$



                                The basic rules for ideal capacitors and inductors are as follows:



                                You can’t change the voltage across an ideal capacitor instantaneously.
                                You can’t change the current through an ideal inductor instantaneously.



                                So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.



                                Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor is



                                $$V_C=fracQC$$



                                Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.



                                The capacitor looks like an open circuit.
                                The inductor looks like a short circuit.



                                That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.



                                Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.



                                In short, your book is correct.



                                Hope this helps.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 2 hours ago









                                Bob DBob D

                                4,3402318




                                4,3402318











                                • $begingroup$
                                  Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                  $endgroup$
                                  – Starboy
                                  2 hours ago
















                                • $begingroup$
                                  Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                  $endgroup$
                                  – Starboy
                                  2 hours ago















                                $begingroup$
                                Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                $endgroup$
                                – Starboy
                                2 hours ago




                                $begingroup$
                                Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached)
                                $endgroup$
                                – Starboy
                                2 hours ago

















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