| The current I drawn from the 5 V source will be [AIEEE 2006] |
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A) 0.33 A
B) 0.5 A
C) 0.67 A
D) 0.17 A
Correct Answer: B
Solution :
| [b] The given circuit can be redrawn as |
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| Which is a balanced condition of Wheatstone's bridge and hence no current flows in the centre resistor, so equivalent circuit would be as shown below. |
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| \[30\,\Omega ||15\,\Omega =10\,\Omega \] \[\left[ \because R=\frac{30\times 15}{30+15} \right]\] |
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| So, current through the circuit, |
| \[I=\frac{V}{R}=\frac{5}{10}=0.5\,A\] |
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