A pure inductor and a pure resistor are connected in series and an ac supply is connected across this combination. Ideal ac voltmeters $V_{1} and V_{2}$ show 120 volt and 160 volt respectively. What is the phase difference between $V_{1} and V_{2}$

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$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{2}$

answer is D.

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Detailed Solution

As current is common through both the elements, so in the inductor, voltage $(V_{1})$ , is ahead of current I, while, in the resistor, voltage $(V_{2})$ is in same phase as current. Hence phase difference between $V_{1} and V_{2}$ is $90^{\circ} .$

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