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(D) For the Wheatstone bridge to be balanced,the ratio of resistances in the arms must be equal: $\frac{R_1}{R_4} = \frac{R_{	ext{bulb}}}{R_3}$.Given

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$25$

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(D) For the Wheatstone bridge to be balanced,the ratio of resistances in the arms must be equal: $\frac{R_1}{R_4} = \frac{R_{	ext{bulb}}}{R_3}$.Given $R_1 = 4 \Omega$,$R_4 = 8 \Omega$,and $R_3 = 12 \Omega$:$\frac{4}{8} = \frac{R_{	ext{bulb}}}{12} \implies R_{	ext{bulb}} = \frac{4 	imes 12}{8} = 6 \Omega$.Using Ohm's Law for the bulb,$V = i R_{	ext{bulb}} = 6i$.Given $V = i(2i + 1)$,we equate the two expressions:$6i = i(2i + 1) \implies 6 = 2i + 1 \implies 2i = 5 \implies i = 2.5 	ext{ A}$.In a balanced Wheatstone bridge,the potential difference across the galvanometer (or bulb) is zero,but here the bulb is part of the arm. The total circuit consists of two parallel branches:Branch $1$: $(4 + 6) \Omega = 10 \Omega$ (This is incorrect based on the diagram; the diagram shows $4 \Omega$ and bulb in series,and $8 \Omega$ and $12 \Omega$ in series).Let's re-evaluate: The bridge is balanced,so the potential at the two nodes connected to the bulb is the same. The total resistance $R_{	ext{eq}}$ is:$R_{	ext{eq}} = \frac{(4 + 6)(8 + 12)}{(4 + 6) + (8 + 12)} = \frac{10 	imes 20}{10 + 20} = \frac{200}{30} = \frac{20}{3} \Omega$.The total current $I = i_1 + i_2$. Since the bridge is balanced,the potential difference across the bulb is $0$ $V$. Wait,if $V=0$,then $i(2i+1)=0 \implies i=0$. Re-reading the problem: The bulb is in the arm. If the bridge is balanced,no current flows through the central branch. Thus $i=0$. However,the question implies the bulb is one of the resistors. If $V=i(2i+1)$ is the characteristic,and $V=iR$,then $R=2i+1$. For balance,$\frac{4}{8} = \frac{R}{12} \implies R=6 \Omega$. $6 = 2i+1 \implies 2i=5 \implies i=2.5 	ext{ A}$.Total voltage $V_b = I_{	ext{total}} 	imes R_{	ext{eq}}$.$I_1 = \frac{V_b}{4+6} = \frac{V_b}{10}$,$I_2 = \frac{V_b}{8+12} = \frac{V_b}{20}$.Since $i=2.5 	ext{ A}$ is the current through the bulb,and in a balanced bridge,the current through the bulb is $0$,there is a contradiction. Assuming the question implies the bulb is in series with the $4 \Omega$ resistor and the bridge is balanced by adjusting $V_b$ such that the potential difference across the bulb is $0$,then $i=0$. Given the options,the intended calculation is $V_b = i(R_1+R_{	ext{bulb}}) = 2.5(4+6) = 25 	ext{ V}$ or similar. Let's use $V_b = I_1(R_1+R_{	ext{bulb}}) = 2.5(10) = 25 	ext{ V}$.

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 $(V_b)$ so that the bridge becomes balanced is (in $V$)

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