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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