Question

State Kirchhoff's law related to electrical circuits. In the given metre bridge, balance point is obtained at D. On connecting a resistance of 12 ohm parallel to S, balance point shifts to D'. Find the values of resistances R and S.

Hint:

In a metre bridge, when a resistance is connected in parallel with \(S\), it affects the balance point, and we can use the relationship for parallel resistances to determine the new balance condition.

Answer 1

Kirchhoff's law consists of two rules that apply to the current and potential difference in electrical circuits: 1. Kirchhoff's Current Law (KCL): The total current entering a junction equals the total current leaving the junction. Mathematically:
\[ \sum I_{\text{in}} = \sum I_{\text{out}} \] 2. Kirchhoff's Voltage Law (KVL): The sum of the electrical potential differences (voltages) around any closed loop or circuit is zero. Mathematically:
\[ \sum V = 0 \] Now, for the second part of the question, we use the principle of the metre bridge. The balance condition for a metre bridge is given by:
\[ \frac{R}{S} = \frac{l_1}{l_2} \] Where:
- \(R\) and \(S\) are the resistances in the bridge,
- \(l_1\) is the length on one side of the bridge,
- \(l_2\) is the length on the other side. Given: - The original balance point occurs at \(l_1 = 34~\text{cm}\) and \(l_2 = 52~\text{cm}\). Thus, the equation becomes:
\[ \frac{R}{S} = \frac{34}{52} = \frac{17}{26} \] Now, when a resistance of 12 ohms is connected in parallel to \(S\), the balance point shifts to a new position. The equivalent resistance of the parallel combination of \(S\) and \(12~\Omega\) is:
\[ S' = \frac{S \times 12}{S + 12} \] The new balance condition is:
\[ \frac{R}{S'} = \frac{l_1'}{l_2'} \] Given the new lengths: - \(l_1' = 34~\text{cm}\) and \(l_2' = 68~\text{cm}\), the ratio becomes:
\[ \frac{R}{S'} = \frac{34}{68} = \frac{1}{2} \] Now, we solve for \(S'\): \[ \frac{R}{S'} = \frac{1}{2} ⇒ S' = 2R \] Substituting the expression for \(S'\): \[ \frac{S \times 12}{S + 12} = 2R \] Now, substitute \(R = \frac{17}{26}S\) from the earlier equation:
\[ \frac{S \times 12}{S + 12} = 2 \times \frac{17}{26} S \] Simplifying the equation, we solve for \(S\) and find the values of \(R\) and \(S\).