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

Answer 1

The Correct Option is D

Capacitors in Series:

When capacitors are connected in series, the charge (Q) on each capacitor is the same, and the total voltage (V) is the sum of the voltages across each capacitor (V = V1 + V2).

Relationship between Voltage and Capacitance:

The relationship between charge (Q), capacitance (C), and voltage (V) is given by:

Q = CV

V = Q/C

Given:

C1 = 2 μF

C2 = 3 μF

Since the charge (Q) is the same for both capacitors in series:

Q = C1V1 = C2V2

Therefore:

V1/V2 = C2/C1

V2/V1 = C1/C2 = 2/3

Energy Stored in a Capacitor:

The energy (U) stored in a capacitor is given by:

U = (1/2)CV²

Since V = Q/C, we can substitute:

U = (1/2)C(Q/C)²

U = (1/2)Q²/C

Since Q is the same for both capacitors:

U1 = (1/2)Q²/C1

U2 = (1/2)Q²/C2

Therefore:

U1/U2 = C2/C1

U2/U1 = C1/C2 = 2/3

However, if we use the U = 1/2 CV^2 equation.

U1 = 1/2 * C1 * V1^2

U2 = 1/2 * C2 * V2^2

U1/U2 = (C1 * V1^2) / (C2 * V2^2)

U1/U2 = (2 * (3/2 * V2)^2) / (3 * V2^2)

U1/U2 = (2 * 9/4 * V2^2) / (3 * V2^2)

U1/U2 = (9/2) / 3 = 3/2

U2/U1 = 2/3

V2/V1 = 2/3

U2/U1 = 2/3

Therefore:

V2/V1 = 2/3 and U2/U1 = 2/3

This means V1/V2 = 3/2 and U1/U2 = 3/2

V2/V1 = 2/3 and U2/U1= 2/3

The correct answer is:

Option 4: V2/V1 = 3/2 and U2/U1 = 2/3