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A uniform conducting wire $\mathrm{AB}$ of length $5 \mathrm{~m}$ and resistance $5 \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \mathrm{~m}$ from $\mathrm{A}$, then the value of $\mathrm{E}$ is
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$3 \mathrm{~V}$
Net resistance of the circuit $=R=5+4+1$ $=10 \Omega$ And resistance due to $3 \mathrm{~m}$ balance length,
$\mathrm{R}^{\prime}=\frac{5}{5} \times 3=3 \Omega$
At balance length, both circuits have equal currents
$I=\frac{E_1}{R}=\frac{E_2}{R^{\prime}} \Rightarrow \frac{10}{10}=\frac{E}{3} \therefore E=3 V$
$\mathrm{R}^{\prime}=\frac{5}{5} \times 3=3 \Omega$
At balance length, both circuits have equal currents
$I=\frac{E_1}{R}=\frac{E_2}{R^{\prime}} \Rightarrow \frac{10}{10}=\frac{E}{3} \therefore E=3 V$
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