Question

Two resistors of resistances 4 $\Omega$ and 8 $\Omega$ are connected in parallel in the left gap of the meter bridge and two resistors of resistances 8 $\Omega$ and 4 $\Omega$ are connected in series in the right gap of the meter bridge. The balancing length from the left end of the bridge wire is

• A. $\frac{100}{11}$ cm
• B. $\frac{200}{9}$ cm
• C. $\frac{180}{11}$ cm
• D. $\frac{200}{11}$ cm

Hint:

- Always simplify parallel and series resistances before using meter bridge formula.
- For left gap in parallel: $R_\text{eq} = \frac{R_1 R_2}{R_1 + R_2}$.
- For right gap in series: $R_\text{eq} = R_1 + R_2$.
- Then apply $l/(100-l) = R_L/R_R$.
- Check that units of length are consistent (cm).

Answer 1

The Correct Option is A

1. Let the length of meter bridge wire = 100 cm. Let balancing length = $l$.
2. Left gap equivalent resistance: $R_L = \frac{4 \cdot 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}~\Omega$
3. Right gap equivalent resistance: $R_R = 8 + 4 = 12~\Omega$
4. Meter bridge formula: $\frac{l}{100 - l} = \frac{R_L}{R_R} = \frac{8/3}{12} = \frac{2}{9}$
5. Solve: $l = \frac{2}{11} \cdot 100 = \frac{100}{11}$ cm