Ncert Exercises & Solutions

An inductor of reactance \(1~\Omega\) and a resistor of \(2~\Omega\) are connected in series to the terminals of a \(6~\text{V}\) (RMS) AC source. The power dissipated in the circuit is:
1. \(8~\text{W}\)
2. \(12~\text{W}\)
3. \(14.4~\text{W}\)
4. \(18~\text{W}\)

Step 1 : Given, X_{L} = 1 Ω, R = 2 Ω

E_{rms} = V, P_{av} = ?

Average power dissipated in the circuit;

P_{av} = E_{rms} I_{rms} cosϕ . . . (i)

Step 2 : I_{rms} = \frac{I_{0}}{\sqrt{2}} = \frac{E_{rms}}{Z}

Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{4 + 1} = \sqrt{5}

I_{rms} = \frac{6}{\sqrt{5}} A

Step 3 : cosϕ = \frac{R}{Z} = \frac{2}{\sqrt{5}}

P_{av} = 6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}} [from Eq . (i)]

= \frac{72}{\sqrt{5} \sqrt{5}} = \frac{72}{5} = 14.4 W

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