The value of current I as shown in the given circuit diagram is

Question

The value of current I as shown in the given circuit diagram is, 2 A, 1 . 5 A, 0 . 9 A, 0 . 2 A

MARKS App · Accepted Answer

The given circuit can be redrawn as

Since, $\frac{R_{A B}}{R_{B C}} = \frac{R_{A D}}{R_{D C}} = \frac{1}{2},$ hence, it is a balanced Wheatstone bridge, therefore current flowing in arm $B D = 0$

$∴ R_{eq} = (10 + 20) ∥ (6 + 12) = 30 ∥ 18$

$= \frac{30 \times 18}{30 + 18} = 11.25 Ω$

$∴ I = \frac{V}{R_{eq}} = \frac{10}{11.25} = 0.89 A = 0.9 A$

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