The value of C is ______. [2021]

Question

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C μ F$ across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take $r m s$ values of the voltages. The magnitude of the phase angle (in degrees) between the current and the supply voltage is $φ$ .

Assume, $π \sqrt{3} \approx 5 .$

Q. The value of C is ______. [2021]

Assume, $π \sqrt{3} \approx 5 .$

Q. The value of C is ______. [2021]

Smart Achievers · Accepted Answer

(100)

From $V_{RMS} = \sqrt{V_{C}^{2} + V_{R}^{2}},$

$\Rightarrow V_{C}^{2} + 100^{2} = 200^{2},$

or, $V_{C}^{2} + 10000 = 40000$

$∴$ $V_{C} = 100 \sqrt{3} V$ $\dots (i)$

$\tan ϕ = \frac{V_{C}}{V_{R}} = \frac{100 \sqrt{3}}{100} = \sqrt{3},$

$∴$ $ϕ = 60 °$ $\dots (i i)$

Power consumed, $P = I_{rms} V_{rms} \cos ϕ = \frac{1}{2} \frac{V_{rms}^{2}}{Z}$

$\Rightarrow 500 = \frac{200^{2}}{Z} \times \frac{1}{2}$

$∴$ $Z = 40 Ω$ $\dots (i i i)$

$\cos ϕ = \frac{R}{Z} \Rightarrow \frac{1}{2} = \frac{R}{40}$

$∴$ $R = 20 Ω$

And, $X_{C} = \sqrt{Z^{2} - R^{2}} = \sqrt{40^{2} - 20^{2}} = 20 \sqrt{3} Ω$

$X_{C} = \frac{1}{C ω} \Rightarrow 20 \sqrt{3} = \frac{1}{C 2 π f}$

$∴$ $C = \frac{1}{2 π f (20 \sqrt{3})} = \frac{1}{20 π \sqrt{3} \times 100} = 10^{- 4} F = 100 μ F$

Solution

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