CBSE Class 12 · 2006 · All India · Alternating CurrentShort Answer

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When an inductor L and a resistor R in series are connected across a 12 V, 50 Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from the applied voltage by

\frac{π}{3}

radian. Calculate the value of R.

Answer: $R = 12 Ω$

Solution

The impedance is

Z = \frac{V _{r m s}}{I _{r m s}} = \frac{12}{0.5} = 24 Ω

For a series LR circuit the power factor is

cos ϕ = \frac{R}{Z}

, so

R = Z cos ϕ = 24 cos \frac{π}{3} = 24 \times \frac{1}{2} = 12 Ω

R = 12 Ω

CBSE Class 12 Board · 2006 · All India · Subjective

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