Question 59: Current Electricity

Question

A galvanometer having a resistance of $$8 \Omega$$ is shunted by a wire of resistance $$2 \Omega$$. If the total current is $$1 \mathrm{~A}$$, the part of it passing through the shunt will be

Accepted Answer

Let total current through the 1 parallel combination be $$I$$, the current through the galv anometer be $$I_g$$ and the current through the shunt be $$I-I_g$$.
The shunted galvanometer is shown in the figure.
The potential difference, $$V_{a b}=\left(V_a-V_bight)$$ is the same for both paths, so $$I_g G=\left(I-I_gight) S$$
$$\Rightarrow I_g(G+S)=I S \Rightarrow \frac{I_g}{I}=\frac{S}{S+G}$$
The fraction of current passing through shunt
$$\begin{aligned}
& =\frac{I-I_g}{I}=1-\frac{I_g}{I}=1-\frac{S}{S+G}=\frac{G}{S+G} \
& =\frac{8}{2+8}=\frac{8}{10}=0.8 \mathrm{~A}
\end{aligned}$$

Answer

0.25 A

Answer

0.8 A

Answer

0.2 A

Answer

0.5 A

Current Electricity

A galvanometer having a resistance of $8 \Omega is shunted by a wire of resistance 2 \Omega. If the total current is 1 \mathrm{~A}$, the part of it passing through the shunt will be

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