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  • The current I drawn from the  source will be: <img alt="" src="https:

The current I drawn from the \mathrm{ 5\: V} source will be:

 

Option: 1

0.33\: \mathrm{A}


Option: 2

0.5\: \mathrm{A}


Option: 3

0.67\: \mathrm{A}


Option: 4

0.17\: \mathrm{A}


Answers (1)

best_answer

The given circuit can be redrawn as

It is a balanced Wheatstone bridge and hence. No current flows in the middle resistor. So, equivalent circuit would be as shown in the figure.

10 \Omega$ and $20 \Omegaresistance are in series, so their equivalent resistance is

\therefore \quad \mathrm{R}^{\prime}=10 \Omega+20 \Omega=30 \Omega

Similarly, 5 \Omega$ and $10 \Omega are in series, so their equivalent resistance is

\mathrm{R}^{\prime \prime}=15 \Omega

\mathrm{R^{\prime} \: and \: R "} are in parallel, so the equivalent resistance of the circuit is

\mathrm{R}=\frac{15 \times 30}{15+30}=10 \Omega \quad \therefore \quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{5 \mathrm{~V}}{10 \Omega}=0.5 \mathrm{~A}

Hence option 2 is correct.

Posted by

Nehul

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