In this figure, the resistance of the coil of galvanometer G is 2. The emf of the cell is 4 V. The ratio of potential difference across C1 and C2 is:

In this figure, the resistance of the coil of galvanometer G is 2… — Solution | JEE Physics

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In this figure, the resistance of the coil of galvanometer $G$ is $2 \, \Omega$ . The emf of the cell is $4 \, \text{V}$ . The ratio of potential difference across $C_1$ and $C_2$ is:

A diamond-shaped circuit with nodes A, B, C, D; resistors 6 ohm on AB and 8 ohm on CD, galvanometer G between B and C, capacitors or cells C1 on AC and C2 on BD, and a 4 volt cell in the outer loop.

A
$\frac{5}{4}$
B
$1$
C
$\frac{4}{5}$
D
$\frac{3}{4}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The resistance of the galvanometer coil is $2 \, \Omega$ and the cell emf is $4 \, \text{V}$ .

Find: The ratio of potential differences across $C_1$ and $C_2$ .

Apply Kirchhoff’s laws. Using the given circuit, calculate the currents and potential drops across $C_1$ and $C_2$ .

The required ratio is:

\frac{V_{C1}}{V_{C2}} = \frac{4}{5}

Therefore, the ratio of potential differences across $C_1$ and $C_2$ is $\frac{4}{5}$ . The correct option is C.

Common mistakes

Using the option label from the solution without checking the numerical value is incorrect because the page marks option A while the computed ratio shown is $\frac{4}{5}$ . Match the value with the listed options, which gives option C.
Ignoring the galvanometer resistance of $2 \, \Omega$ is incorrect because it affects the current distribution in the bridge network. Include the galvanometer branch while applying Kirchhoff’s laws.
Taking the ratio of resistances directly as the ratio of potential differences is incorrect because the circuit is not a simple series divider. First determine branch currents or potential drops using circuit laws, then form the ratio $V_{C1}/V_{C2}$ .

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Answer

Step-by-step solution

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