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प्रश्न
A capacitor C1 of capacitance 1 μF and a capacitor C2 of capacitance 2 μF are separately charged by a common battery for a long time. The two capacitors are then separately discharged through equal resistors. Both the discharge circuits are connected at t = 0.
(a) The current in each of the two discharging circuits is zero at t = 0.
(b) The currents in the two discharging circuits at t = 0 are equal but not zero.
(c) The currents in the two discharging circuits at t = 0 are unequal.
(d) C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge.
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उत्तर
(b) The currents in the two discharging circuits at t = 0 are equal but not zero.
(d) C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge.
Let the voltage of the battery connected to the capacitors be V. Both the capacitors will charge up to the same potential (V).
The charge on the capacitors C1 is C1V = (1 μF)×V
The charge on the capacitors C2 is C2V = (2 μF)×V
The charge on the discharging circuit at an instant t,
\[Q = CV e^{- t/RC}\]
The current through the discharging circuit,
\[\frac{dQ}{dt} = - \frac{CV}{RC} e^{- t/RC} = \frac{V}{R} e^{- t/RC}\]
At t = 0, the current through the discharging circuit will be `V/R` for both the capacitors.
Let the time taken by the capacitor C1 to lose 50% of the charge be t1.
\[Q_1 = \frac{C_1 V}{2}\]
\[\frac{C_1 V}{2} = C_1 V e^{- t_1 /RC} \]
\[\frac{1}{2} = e^{- t_1 /RC}\]
Taking natural log on both sides, we get:-
\[- \ln2 = - \frac{t_1}{R C_1}\]
\[ t_1 = R C_1 \ln2\]
Similarly,
Time taken for capacitor C2:-
\[t_2 = R C_2 \ln2\]
As, C1 < C2, t1 < t2
Thus, we can say that C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge.
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