Switch S is closed at $ t = 0 $ , $ {I_{10}} $ is the current supplied by the battery just after closing the switch S. $ {Q_1} $ , $ {Q_2} $ and $ {Q_3} $ are the charges on the capacitors of $ 10\mu F $ , $ 20\mu F $ and $ 30\mu F $ in steady state respectively. $ {I_{20}} $ is the current supplied by the battery in the circuit at steady state. Choose the correct statement(s).
(A) $ {I_{10}} > {I_{20}} $
(B) $ {I_{10}} < {I_{20}} $
(C) $ {Q_1} < {Q_2} < {Q_3} $
(D) $ {Q_1} < {Q_3} < {Q_2} $
Answer
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Hint: No current was flowing through the capacitors before the switch was closed at $ t = 0 $ . We know that current is the flow of charge.
Formula Used: The formulae used in the solution are given here.
Current $ I = C\dfrac{{dV}}{{dt}} $ , where $ C $ the capacitance in Farad and voltage across the capacitor is $ V $ .
Charge on the capacitor, $ Q = CV $ .
Complete Step by Step Solution
No flow of charge to and from the capacitors plates was zero, prior to $ t = 0 $ . Hence as charge on capacitor plates varies proportionally with the voltage across them, therefore voltage across them is zero before $ t = 0 $ . We can say that the capacitor was short circuited or it behaved as if it was being short circuited.
We know, current through capacitor, $ I $ and voltage across the capacitor, $ V $ , is related as
$ I = C\dfrac{{dV}}{{dt}} $ , where $ C $ is the capacitance in Farad.
Following from the above, when the switch is turned on at $ t = 0 $ , the capacitor behaves as being short circuited , for the initial time being.
The circuit thus looks like,
On solving, we see that the resistances $ 10\Omega $ , $ 6\Omega $ , $ 15\Omega $ , $ 6\Omega $ are in parallel. So equivalent resistance,
$ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{10}} + \dfrac{1}{6} + \dfrac{1}{{15}} + \dfrac{1}{6} $
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{2} $ .
So equivalent resistance, $ {R_{eq}} = 2\Omega $ .
There is another resistance $ 2\Omega $ connected in series to the arrangement for which $ {R_{eq}} = 2\Omega $ .
Thus, resistance $ R = 4\Omega $ .
By Ohm’s Law, $ V = IR $ .
Thus, $ {I_{10}} = \dfrac{{60}}{4} = 15A $ .
At steady state, that is, after some time, the capacitor starts behaving like an open circuit.
Thus two $ 6\Omega $ resistors are in parallel with two $ 12\Omega $ resistors.
Equivalent resistance is thus,
$ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{2 \times 6}} + \dfrac{1}{{2 \times 12}} $
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{8} $
S, equivalent resistance $ {R_{eq}} = 8\Omega $ .
There are two resistances $ 10\Omega $ and $ 2\Omega $ connected in series to the arrangement for which $ {R_{eq}} = 8\Omega $
By Ohm’s Law, $ V = IR $ .
At $ {I_{20}} = \dfrac{{60}}{{20}} = 3A. $
So $ {I_{20}} > {I_{10}} $ .
Hence Option A is correct.
Now charge on the capacitor, $ Q = CV $ .
Voltage across the $ 10\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_1} = 10\mu F \times 30 $ .
Voltage across the $ 20\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_2} = 20\mu F \times 42 $ .
Voltage across the $ 30\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_3} = 30\mu F \times 24 $ .
It is quite apparent, $ {Q_3} > {Q_2} > {Q_1} $ .
Hence Option D is also correct.
Note:
We can conclude that if voltage changes abruptly, current through the capacitor is spiked up. So, the tendency of the capacitor is to avoid the abrupt change in voltage across it.
Formula Used: The formulae used in the solution are given here.
Current $ I = C\dfrac{{dV}}{{dt}} $ , where $ C $ the capacitance in Farad and voltage across the capacitor is $ V $ .
Charge on the capacitor, $ Q = CV $ .
Complete Step by Step Solution
No flow of charge to and from the capacitors plates was zero, prior to $ t = 0 $ . Hence as charge on capacitor plates varies proportionally with the voltage across them, therefore voltage across them is zero before $ t = 0 $ . We can say that the capacitor was short circuited or it behaved as if it was being short circuited.
We know, current through capacitor, $ I $ and voltage across the capacitor, $ V $ , is related as
$ I = C\dfrac{{dV}}{{dt}} $ , where $ C $ is the capacitance in Farad.
Following from the above, when the switch is turned on at $ t = 0 $ , the capacitor behaves as being short circuited , for the initial time being.
The circuit thus looks like,
On solving, we see that the resistances $ 10\Omega $ , $ 6\Omega $ , $ 15\Omega $ , $ 6\Omega $ are in parallel. So equivalent resistance,
$ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{10}} + \dfrac{1}{6} + \dfrac{1}{{15}} + \dfrac{1}{6} $
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{2} $ .
So equivalent resistance, $ {R_{eq}} = 2\Omega $ .
There is another resistance $ 2\Omega $ connected in series to the arrangement for which $ {R_{eq}} = 2\Omega $ .
Thus, resistance $ R = 4\Omega $ .
By Ohm’s Law, $ V = IR $ .
Thus, $ {I_{10}} = \dfrac{{60}}{4} = 15A $ .
At steady state, that is, after some time, the capacitor starts behaving like an open circuit.
Thus two $ 6\Omega $ resistors are in parallel with two $ 12\Omega $ resistors.
Equivalent resistance is thus,
$ \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{2 \times 6}} + \dfrac{1}{{2 \times 12}} $
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{8} $
S, equivalent resistance $ {R_{eq}} = 8\Omega $ .
There are two resistances $ 10\Omega $ and $ 2\Omega $ connected in series to the arrangement for which $ {R_{eq}} = 8\Omega $
By Ohm’s Law, $ V = IR $ .
At $ {I_{20}} = \dfrac{{60}}{{20}} = 3A. $
So $ {I_{20}} > {I_{10}} $ .
Hence Option A is correct.
Now charge on the capacitor, $ Q = CV $ .
Voltage across the $ 10\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_1} = 10\mu F \times 30 $ .
Voltage across the $ 20\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_2} = 20\mu F \times 42 $ .
Voltage across the $ 30\mu F $ capacitor at steady state= $ 30V $ .
$ {Q_3} = 30\mu F \times 24 $ .
It is quite apparent, $ {Q_3} > {Q_2} > {Q_1} $ .
Hence Option D is also correct.
Note:
We can conclude that if voltage changes abruptly, current through the capacitor is spiked up. So, the tendency of the capacitor is to avoid the abrupt change in voltage across it.
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