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

$Step 1 : Given, X_{L} = 1 Ω, R = 2 Ω$

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

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

Average power dissipated in the circuit;

$Average power dissipated in the circuit;$

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

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

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

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

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

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

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

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

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

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

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

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

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

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