The potential difference \( V \) across the filament of the bulb shown in the given Wheatstone bridge varies as \( V = i(2i + 1) \), where \( i \) is the current in ampere through the filament of the bulb. The emf of the battery (\( V_a \)) so that the bridge becomes balanced is:
Show Hint
In a balanced Wheatstone bridge, the ratio of resistances satisfies \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \).
- Use \( V = IR \) to find the required potential difference.
The Wheatstone bridge is balanced when the ratio of resistances in the two arms is equal:
\[
\frac{R_1}{R_2} = \frac{R_3}{R_4}
\]
Given resistances:
- \( R_1 = 8\Omega \), \( R_2 = 12\Omega \)
- \( R_3 = R_{\text{bulb}} \), \( R_4 = 4\Omega \)
1. Find Resistance of the Bulb \( R_{\text{bulb}} \):
Since the bridge is balanced,
\[
\frac{8}{12} = \frac{R_{\text{bulb}}}{4}
\]
\[
R_{\text{bulb}} = \frac{8}{12} \times 4 = \frac{32}{12} = \frac{8}{3} \Omega
\]
2. Find Current through the Bulb \( i \):
The voltage-current relation is given by:
\[
V = i(2i + 1)
\]
Using \( V = iR_{\text{bulb}} \):
\[
i \left( \frac{8}{3} \right) = i(2i + 1)
\]
Equating,
\[
\frac{8}{3} = 2i + 1
\]
\[
2i = \frac{8}{3} - 1 = \frac{5}{3}
\]
\[
i = \frac{5}{6} A
\]
3. Find EMF of Battery (\( V_a \)):
The total resistance in the balanced bridge circuit:
\[
R_{\text{eq}} = 8 + 12 = 20 \Omega
\]
The voltage supplied is:
\[
V_a = i R_{\text{eq}} = \frac{5}{6} \times 30
\]
\[
= 25 V
\]
Thus, the correct answer is \(\boxed{25 V}\).
