Question

If the galvanometer G does not show any deflection in the circuit shown, the value of R is given by:

• A. 400Ω
• B. 200Ω
• C. 50Ω
• D. 100Ω

Hint:

Applying Kirchhoff's law in the circuit. 

Answer 1

The Correct Option is D

To solve the problem, we need to determine the value of resistor \(R\) such that the galvanometer \(G\) shows no deflection. This indicates a balanced condition in the circuit.

For a galvanometer to show no deflection, the potential difference across it must be zero. This implies that the potential at both ends of the galvanometer is equal.

In the given circuit, this condition can be understood using the Wheatstone bridge principle, even though it is not a conventional bridge. Let's apply Kirchhoff's Voltage Law (KVL) and the concepts of potential difference to find the resistance \(R\).

1. The current through the 400Ω resistor can be calculated using Ohm's law as it is connected to a 10V battery. Let this current be \(I_1\): 
   \(I_1 = \frac{10 \text{V}}{400 \Omega} = \frac{1}{40} \, \text{A}\).
2. The potential difference across the 400Ω resistor is \(4 \, \text{V}\) (since \(400 \times \frac{1}{40} = 10\) gives \(6 \, \text{V}\) drop for the remainder).
3. This means the potential at the point before the resistor \(R\) is \(6 \, \text{V}\) higher than its other side connected to the 2V source.
4. The potential on the other branch, parallel to \(R\), must therefore also differ by \(6 \, \text{V}\) to ensure no current through \(G\), balancing the 2V source.
5. The entire loop gives: \(10 \, \text{V} - 6 \, \text{V} = 4 \, \text{V} = 2 \, \text{V} + 2 \, \text{V} \Rightarrow R = 100 \, \Omega\). This satisfies \(I_1 \cdot R = 4 \, \text{V}\) with \(R = 100 \, \Omega\).

Thus, the value of resistance \(R\) which would result in no deflection of the galvanometer is 100Ω.