SATHEE: alternating-currents Question 33

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alternating-currents Question 33

Question: Q. 1. An inductor of reactance $1 Ω$ and a resistor of $2 Ω$ are connected in series to the terminals of a $6 V (r m s)$ AC source. The power dissipated in the circuit is (a) $8 W$ . (b) $12 W$ . (c) $14.4 W$ . (d) $18 W$ .

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Solution:

Ans. Correct option : (c)

Explanation :

Here, $E_{rms} = 6 V, X_{L} = 1 Ω, R = 2 Ω, p$

We know that, average power dissipated in the circuit is given by,

$\begin{aligned} P_{a v} & = E_{r m s} I_{r m s} \cos ϕ I_{r m s} & = \frac{I_{0}}{\sqrt{2}} = \frac{E_{sms}}{Z} Z & = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{4 + 1} = \sqrt{5} Ω ∴ I_{r m s} & = \frac{6}{\sqrt{5}} A \end{aligned}$

$\begin{aligned} \cos ϕ = \frac{R}{Z} = \frac{2}{\sqrt{5}} & P_{a v} = 6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}} = 14.4 W \end{aligned}$

Alternating Currents

alternating-currents Question 33

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Alternating Currents

alternating-currents Question 33

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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