Time constant of a C-R circuit is `2/(ln(2))` second. Capacitor is discharged at time `t=0`. The ratio of charge on the capacitor at time `t=2s` and `t=6s` is

Options

A. `3:1`
B. `8:1`
C. `4:1`
D. `2:1`

Answer

To find the ratio of charge on the capacitor at time \( t = 2s \) and \( t = 6s \) in a C-R circuit, we can use the formula for the charge \( Q(t) \) on a discharging capacitor, which is given by:

\[ Q(t) = Q_0 e^{-t/\tau} \]

where:
- \( Q_0 \) is the initial charge on the capacitor,
- \( \tau \) is the time constant of the circuit,
- \( t \) is the time in seconds.

Given that the time constant \( \tau = \frac{2}{\ln(2)} \) seconds, we can calculate the charges at \( t = 2s \) and \( t = 6s \).

### Step 1: Calculate the charge at \( t = 2s \)

Using the formula:

\[ Q(2) = Q_0 e^{-2/\tau} \]

Substituting \( \tau = \frac{2}{\ln(2)} \):

\[ Q(2) = Q_0 e^{-2 / \left(\frac{2}{\ln(2)}\right)} = Q_0 e^{-\ln(2)} = Q_0 \cdot \frac{1}{2} = \frac{Q_0}{2} \]

### Step 2: Calculate the charge at \( t = 6s \)

Using the same formula:

\[ Q(6) = Q_0 e^{-6/\tau} \]

Substituting \( \tau = \frac{2}{\ln(2)} \):

\[ Q(6) = Q_0 e^{-6 / \left(\frac{2}{\ln(2)}\right)} = Q_0 e^{-3 \ln(2)} = Q_0 \cdot (e^{\ln(2)})^{-3} = Q_0 \cdot \left(\frac{1}{2}\right)^3 = Q_0 \cdot \frac{1}{8} = \frac{Q_0}{8} \]

### Step 3: Calculate the ratio of charges

Now we can find the ratio of the charges at \( t = 2s \) and \( t = 6s \):

\[ \text{Ratio} = \frac{Q(2)}{Q(6)} = \frac{\frac{Q_0}{2}}{\frac{Q_0}{8}} = \frac{Q_0}{2} \cdot \frac{8}{Q_0} = \frac{8}{2} = 4 \]

Thus, the ratio of charge on the capacitor at \( t = 2s \) and \( t = 6s \) is \( 4:1 \).

### Final Answer:
The ratio of charge on the capacitor at \( t = 2s \) and \( t = 6s \) is \( 4:1 \).

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