In the following circuit, the current through the resistor $R = 2 Ω$ is $I$ amperes. The value of $I$ is [2015]

Question

Smart Achievers · Accepted Answer

(1)

In the fig. (a), no current flows through the $8 Ω$ resistance.

Again, the equivalent resistance of the balanced Wheatstone bridge. In Fig. (b), no current flows through the $10 Ω$ resistance.

$∴$ $R_{eq} = \frac{6 \times 18}{24} = \frac{9}{2} Ω$

Therefore, the current through the resistor $R = 2 Ω$ is

$I = \frac{V}{R} = \frac{6.5}{2 + \frac{9}{2}} = \frac{6.5}{6.5} = 1 A$

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