Question

Two parallel plate capacitors of capacitances 2 \(\mu F\) and 3 \(\mu F\) are joined in series and the combination is connected to a battery of V volts. The values of potential across the two capacitors \(V_1\) and \(V_2\) and energy stored in the two capacitors \(U_1\)and \(U_2\) respectively are related as_____
Fill in the blank with the correct answer from the options given below

• A. \(\frac{V_2}{V_1} = \frac{3}{2} \quad \text{and} \quad \frac{U_2}{U_1} = \frac{2}{3}\)
• B. \(\quad \frac{V_2}{V_1} = \frac{U_2}{U_1} = \frac{3}{2}\)
• C. \(\quad \frac{V_2}{V_1} = \frac{2}{3} \quad \text{and} \quad \frac{U_2}{U_1} = \frac{3}{2}\)
• D. \(\quad \frac{V_2}{V_1} = \frac{U_2}{U_1} = \frac{2}{3}\)

Answer 1

The Correct Option is D

The objective is to determine the ratios \(\frac{V_2}{V_1}\) and \(\frac{U_2}{U_1}\) for two capacitors, with capacitances 2 \(\mu F\) and 3 \(\mu F\), connected in series to a V-volt battery.

The equivalent capacitance \(C_t\) for capacitors in series is calculated as:

\(\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2}\)

Substituting the given capacitance values yields:

\(\frac{1}{C_t} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\)

Therefore, the total capacitance is:

\(C_t = \frac{6}{5} \mu F\)

In a series connection, the charge \(Q\) is uniform across all capacitors. The voltage across each capacitor is determined by \(Q=CV\):

\(V_1 = \frac{Q}{C_1} \), \( V_2 = \frac{Q}{C_2}\)

Consequently, the voltage ratio is:

\(\frac{V_2}{V_1} = \frac{C_1}{C_2} = \frac{2}{3}\).
The energy stored in a capacitor is given by:

\(U = \frac{1}{2}CV^2\)

The ratio of energies stored is thus:

\(\frac{U_2}{U_1} = \frac{\frac{1}{2}C_2V_2^2}{\frac{1}{2}C_1V_1^2} = \frac{C_2V_2^2}{C_1V_1^2} = \frac{\frac{Q^2}{C_2}}{\frac{Q^2}{C_1}} = \frac{C_1}{C_2} = \frac{2}{3}\).
Both ratios, \(\frac{V_2}{V_1}\) and \(\frac{U_2}{U_1}\), are equal to \(\frac{2}{3}\), confirming the result:

\(\frac{V_2}{V_1} = \frac{U_2}{U_1} = \frac{2}{3}\)