Ncert Exercises & Solutions

An electrical device draws 2 kW power from AC mains (voltage 223 V (RMS)= $\sqrt{50000}$ V). The current differs (lags) in phase by $ϕ (tanϕ = \frac{- 3}{4})$ as compared to voltage. Find (a) R, (b) $X_{C} - X_{L}$ (c) $I_{M}$ . Another device has twice the values for $R, X_{C} and X_{L}$ . How are the answers affected?

Hint: Use the formula of impedance.

Step 1: Given, power drawn=P =2kW=2000 W

$tanϕ = - \frac{3}{4}, I_{M} = I_{0} = ?, R = ?, X_{C} - X_{L} = ?$

$V_{rms} = V = 223 V$

Power

$P = \frac{V^{2}}{Z}$

$\Rightarrow Z = \frac{V^{2}}{P} = \frac{223 \times 223}{2 \times 10^{3}} = 25 Ω$

Impedance,

$Z = 25 Ω$

Step 2: Impedance,

$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$

$\Rightarrow 25 = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$

$or 625 = R^{2} + {(X_{L} - X_{C})}^{2} . . . (i)$

Again,

$\tan ϕ = \frac{X_{L} - X_{C}}{R} = \frac{3}{4}$

$X_{L} - X_{C} = \frac{3 R}{4} . . . (i i)$

From Eq. (ii), we put

$X_{L} - X_{c} = \frac{3 R}{4}$ in Eq. (i), we get;

$625 = R^{2} + {(\frac{3 R}{4})}^{2} = R^{2} + \frac{9 R^{2}}{16}$

$625 = \frac{25 R^{2}}{16}$

(a) Resistance

$R = \sqrt{25 \times 16} = \sqrt{400} = 20 Ω$

(b)

$X_{L} - X_{C} = \frac{3 R}{4} = \frac{3}{4} \times 20 = 15 Ω$

$I_{M} = \sqrt{2} I = \sqrt{2} \frac{V}{Z} = \frac{223}{25} \times \sqrt{2} = 12.6 A$

As R,

$X_{C}, X_{L}$ are all doubled,

$\tan ϕ$ does not change, Z is doubled, the current is halved. So, power is also halved.

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