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In the following network, the current through galvanometer will
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Verified Answer
The correct answer is:
flow from $\mathrm{S}$ to $\mathrm{Q}$
Consider the following diagram:
As the bridge is not balanced so the current will flow through wire.
Consider the loop PSQP:
$$
-(I-i)-i_g G+4 i=0
$$
$\Rightarrow-I-i_g G+5 i=0 \quad---(1)$
Now, consider the loop SQRS:
$$
\begin{aligned}
& -i_g G-4\left(i+i_g\right)+3\left(I-i-i_g\right)=0 \\
& \Rightarrow+3 I-i_g(G+7)-7 i=0 \quad--(2)
\end{aligned}
$$
On considering following operation, $7 \times \mathrm{eq}^{\mathrm{n}}(1)+5 \times \mathrm{eq}^{\mathrm{n}}(2)$
$\begin{aligned} & 4 I-i_g(12 G-35)=0 \\ & \Rightarrow i_g=\frac{4 I}{(12 G+35)}\end{aligned}$
Therefore, the current will flow through from $\mathrm{S}$ to $\mathrm{Q}$, as the value of $I$ and hence, the value of $i_g$ will be positive. As, $\mathrm{G}$ is also positive, it is the resistance of the galvanometer.
As the bridge is not balanced so the current will flow through wire.
Consider the loop PSQP:
$$
-(I-i)-i_g G+4 i=0
$$
$\Rightarrow-I-i_g G+5 i=0 \quad---(1)$
Now, consider the loop SQRS:
$$
\begin{aligned}
& -i_g G-4\left(i+i_g\right)+3\left(I-i-i_g\right)=0 \\
& \Rightarrow+3 I-i_g(G+7)-7 i=0 \quad--(2)
\end{aligned}
$$
On considering following operation, $7 \times \mathrm{eq}^{\mathrm{n}}(1)+5 \times \mathrm{eq}^{\mathrm{n}}(2)$
$\begin{aligned} & 4 I-i_g(12 G-35)=0 \\ & \Rightarrow i_g=\frac{4 I}{(12 G+35)}\end{aligned}$
Therefore, the current will flow through from $\mathrm{S}$ to $\mathrm{Q}$, as the value of $I$ and hence, the value of $i_g$ will be positive. As, $\mathrm{G}$ is also positive, it is the resistance of the galvanometer.
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