Assuming the voltage is kept constant, what is the ratio between the new current and the old current when a third identical resistor is added in parallel with the earlier two?

\frac{4}{9}

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\frac{2}{3}

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\frac{3}{2}

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\frac{9}{4}

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Solution

The correct option is C $\frac{3}{2}$
When the two identical resistors are in parallel, the equivalent resistance, $R_{1} = \frac{R R}{R + R} = R / 2$
When a third identical resistor is added to them in parallel, the equivalent resistance, $R_{2} = (1 / R + 1 / R + 1 / R)^{- 1} = R / 3$
As voltage V is kept constant so by Ohm's law, $V = I_{1} R_{1}$ and $V = I_{2} R_{2}$
Thus, $\frac{I_{2}}{I_{1}} = \frac{R_{1}}{R_{2}} = \frac{R / 2}{R / 3} = 3 / 2$

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