Question

A capacitor is discharging through a resistor R. Consider in time t₁, the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored reduces to one eighth of its initial value. The ratio t₁/t₂ will be

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{6}\)

Answer 1

The Correct Option is D

To solve this problem, we need to consider the discharge behavior of a capacitor through a resistor. The energy stored in a capacitor and the charge stored decay exponentially over time when the capacitor discharges through a resistor.

The energy stored in a capacitor is given by:

E = \frac{1}{2}C V^2

where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

When a capacitor discharges, the energy decreases as:

E(t) = E_0 e^{-2t/RC}

E_0 is the initial energy.

The charge on the capacitor at any time is given by:

Q(t) = Q_0 e^{-t/RC}

Q_0 is the initial charge.

Let's consider:

• In time t_1, the energy reduced to half: E(t_1) = \frac{1}{2}E_0
• In time t_2, the charge reduced to one eighth: Q(t_2) = \frac{1}{8}Q_0

Energy Reduction Analysis:

Using the energy decay equation:

\frac{1}{2}E_0 = E_0 e^{-2t_1/RC}

\Rightarrow e^{-2t_1/RC} = \frac{1}{2}

Taking logarithm on both sides:

-\frac{2t_1}{RC} = \ln \left(\frac{1}{2}\right)

t_1 = \frac{RC}{2} \ln(2)

Charge Reduction Analysis:

Using the charge decay equation:

\frac{1}{8}Q_0 = Q_0 e^{-t_2/RC}

\Rightarrow e^{-t_2/RC} = \frac{1}{8}

Taking logarithm on both sides:

-\frac{t_2}{RC} = \ln \left(\frac{1}{8}\right)

t_2 = RC \ln(8)

Calculating the Ratio:

Now, we find the ratio \frac{t_1}{t_2}:

\frac{t_1}{t_2} = \frac{\frac{RC}{2}\ln(2)}{RC\ln(8)}

\Rightarrow \frac{t_1}{t_2} = \frac{1}{2} \times \frac{\ln(2)}{\ln(8)}

Since \ln(8) = \ln(2^3) = 3\ln(2), we substitute:

\frac{t_1}{t_2} = \frac{1}{2} \times \frac{\ln(2)}{3\ln(2)}

\Rightarrow \frac{t_1}{t_2} = \frac{1}{6}

Thus, the ratio \frac{t_1}{t_2} is \frac{1}{6}, which corresponds to the correct answer.