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In the given circuit, the voltmeter reading is 4.5 V. Assuming that the voltmeter is ideal, current through 12\ \Omega resistance is -

Option: 1

1 A


Option: 2

 0.5 A
 


Option: 3

0.25 A


Option: 4

0.1 A


Answers (1)

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The voltmeter is ideal, its resistance R_{v}\rightarrow \infty Fig. shows the current distribution in the circuit . Voltmeter will not draw any current.

Potential difference across 9 \Omega  resistance=4.5 \mathrm{~V} (given) Hence, current in 9 \Omega resistance
\begin{aligned} & =\frac{4.5}{9}=0.5 \mathrm{~A}\left(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\right)\\ & i.e., \mathrm{r}_1^{\prime}=0.5 \mathrm{~A} \end{aligned}

The same current \left(\mathrm{I}_1^{\prime}\right) passes through 3 \Omega. Obviously, 9 \Omega and 3 \Omega are in series and their equivalent, i.e., 12 \Omega is in parallel with 6 \Omega between A and B. Dividing the current in the inverse ratio of resistances between A and B,
\begin{aligned} & \quad \frac{\mathrm{I}_1^{\prime}}{I_1^{\prime \prime}}=\frac{6}{12}=\frac{1}{2} \\ & \mathrm{I}_1^{\prime \prime}=2 \mathrm{I}_1^{\prime}=2 \times 0.5=1 \mathrm{~A} \\ & \text { and } \mathrm{I}_1=\mathrm{I}_1^{\prime \prime}+\mathrm{I}_1^{\prime}=0.5+1=1.5 \mathrm{~A} \end{aligned}
at junction \mathrm{C}, \mathrm{I}_2 divides into three parts. Since the resistances 10 \Omega, 12 \Omega, 15 \Omega are in parallel between C and D, current will distribute in the inverse ratio of resistances.
\begin{aligned} & \therefore \mathrm{I}_2^{\prime}: \mathrm{I}_2^{\prime \prime}: \mathrm{I}_2^{\prime \prime \prime}=\frac{1}{10}: \frac{1}{12}: \frac{1}{15} \\ &=6: 5: 4 \\ & \mathrm{I}_2^{\prime}=\frac{5}{15} \times 1.5=0.5 \mathrm{amp} \end{aligned}

So \mathrm{I}_2^{\prime}=6 \mathrm{k}, \mathrm{I}_2^{\prime \prime}=5 \mathrm{k}, \mathrm{I}_2^{ \prime \prime \prime}=4 \mathrm{k}
( \mathrm{k} being a constant of proportionality)
and \mathrm{I}_1=\mathrm{I}_2+\mathrm{I}_2^{\prime \prime}+\mathrm{I}_2^{\prime \prime \prime}=15 \mathrm{k}
\begin{aligned} & \text { but } \mathrm{l}_1=1.5 \mathrm{~A} \\ & \therefore 15 \mathrm{k}=1.5 \\ & \text { or } \mathrm{k}=0.1 \\ & =5 \mathrm{k}=0.5 \mathrm{~A} \end{aligned}
so i_2^{\prime \prime}=5 \mathrm{k}=0.5 \mathrm{~A}
Thus current through 12 \Omega resistance is 0.5 \mathrm{~A}

 

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