Two identical capacitors $A$ and $B$, charged to the same potential $5V$ are connected in two different circuits as shown below at time $t = 0$. If the charge on capacitors $A$ and $B$ at time $t = CR$ is $Q_A$ and $Q_B$ respectively, then (Here e is the base of natural logarithm)
- $Q_{A}=VC, Q_{B}=CV$
- $Q_{A}=\frac{CV}{2}, Q_{B}=\frac{VC}{e}$
- $Q_{A}=\frac{VC}{e}, Q_{B}=\frac{CV}{2}$
- $Q_{A}=VC, Q_{B}=\frac{VC}{e}$
The Correct Option is D
Solution and Explanation
To solve this problem, we need to understand the behavior of capacitors in circuits over time, specifically when they are discharging.
Given that capacitors \( A \) and \( B \) are initially charged to a potential \( V = 5 \, \text{V} \) and connected in two different circuits at time \( t = 0 \), we need to calculate the charge on each capacitor at time \( t = CR \).
- For Capacitor \( A \):
- We assume capacitor \( A \) is in an open circuit or connected in such a way that it does not discharge. Therefore, the charge \( Q_A \) on capacitor \( A \) remains constant at its initial value, which is given by the formula:
- \(Q_A = VC\)
- Thus, at any time \( t \), including \( t = CR \), the charge remains \(Q_A = VC\).
- For Capacitor \( B \):
- We assume capacitor \( B \) is connected in a simple discharging RC circuit. In such a circuit, the charge on a capacitor as a function of time \( t \) is given by:
- \(Q(t) = Q_0 e^{-t/RC}\)
- Where \( Q_0 \) is the initial charge, \( R \) is the resistance, \( C \) is the capacitance, and \( e \) is the base of natural logarithms.
- Initially, the charge \( Q_0 \) is \( VC \).
- At time \( t = RC \), the charge on the capacitor \( B \) becomes:
- \(Q_B = VC \cdot e^{-1} = \frac{VC}{e}\)
Thus, at time \( t = CR \), we have:
- \(Q_{A} = VC\)
- \(Q_{B} = \frac{VC}{e}\)
Therefore, the correct answer is:
- \(Q_{A} = VC, Q_{B} = \frac{VC}{e}\)
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