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Three $10 Ω, 2 W$ resistors are connected as in figure. The maximum possible voltage between points A and B without exceeding the power dissipation limits of any of the resistors is

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$5 \sqrt{3} V$

$3 \sqrt{5} V$

15 V

$\frac{5}{3} V$

answer is B.

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Detailed Solution

Power = $\frac{V^{2}}{R}$

or $V = \sqrt{P R} = \sqrt{2 \times 10} = \sqrt{20} = 2 \sqrt{5} V$

Clearly, voltage across single resistor of $10 Ω$ cannot exceed $2 \sqrt{5} V .$ Note that the resistance of parallel combination is half of $10 Ω .$ Thus, the maximum possible voltage between A and B is $3 \sqrt{5} V .$

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