In the circuit shown the cells A and B have negligible resistances. For $V_A =12\, V,\, R_1 =500\, \Omega $ and $ R =100\, \Omega $ the galvanometer (G) shows no deflection. The value of $V_B$ is

Question

In a meter bridge experiment, a resistance of $ 10 \, \Omega $ is balanced by a resistance $ X $ with a balance point at $ 40 \, \mathrm{cm} $. What is the value of $ X $?

Answer 1

To determine the value of $ V_B $ in the given circuit where the galvanometer shows no deflection, we need to understand that this condition indicates a balanced Wheatstone bridge. In a balanced Wheatstone bridge, the ratio of the resistances in one arm is equal to the ratio in the other arm.

Given:

$ V_A = 12\, V $
$ R_1 = 500\, \Omega $
$ R = 100\, \Omega $

In the bridge, since the galvanometer shows no deflection, the potential difference across it is zero. This implies that the potential difference across resistor $ R_1 $ connected with $ V_A $ is equal to the potential difference across resistor $ R $ and $ V_B $.

The formula for balancing a Wheatstone bridge is:

$\frac{V_A}{R_1} = \frac{V_B}{R}$

Substituting the given values:

$\frac{12}{500} = \frac{V_B}{100}$

Simplifying the equation:

12 \times 100 = V_B \times 500

1200 = 500 \times V_B

V_B = \frac{1200}{500}

V_B = 2.4 \, V

Given the options, the closest value to $ V_B $ is $ 2 \, V $.

Thus, the value of $ V_B $ is 2 V.

The correct answer is 2 V, as no deflection in the galvanometer indicates a balanced bridge condition, which has been satisfied by the derived calculations.

Answer 2