Comprehension
In the circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C\, \mu F$ 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 supply voltage is $\phi $. Assume, $\pi \sqrt{3} \approx 5$
Question: 1
The value of C is ___.
The value of C is ___.
Updated On: May 25, 2024
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Correct Answer: 100
Solution and Explanation
\(P = \frac{V^2}{2}\)
\(⇒\) \(500 = \frac{100^2}{R}\)
\(⇒ R = 20 Ω\)
Now across resistance 500 = I × 100
\(⇒\) I rms = 5 A
Vrms = 200 V,
\(\frac{V_{\text{rms}}}{\text{real}} = 100 \, \text{V}\)
\(\cos \varphi = \frac{100}{200} = \frac{1}{2} \quad \Rightarrow \quad \varphi = 60^\circ\)
\(\tan \varphi = \frac{X_C}{R} = \frac{1}{\omega RC}\)
\(\sqrt{3} = \frac{1}{100\pi(20)C}\)
\(C = \frac{1}{{20\pi\sqrt{3} \times 100}}\)
\(C = 10^{-4} \, \text{F}\)
\(= 100 μF\)
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Question: 2
The value of 𝜑 is ___.
The value of 𝜑 is ___.
Updated On: May 23, 2024
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Correct Answer: 60
Solution and Explanation
\(P = \frac{V^2}{2}\)
\(⇒\) \(500 = \frac{100^2}{R}\)
\(⇒ R = 20 Ω\)
Now across resistance 500 = I × 100
\(⇒\) I rms = 5 A
Vrms = 200 V,
\(\frac{V_{\text{rms}}}{\text{real}} = 100 \, \text{V}\)
\(\cos \varphi = \frac{100}{200} = \frac{1}{2} \quad \Rightarrow \quad \varphi = 60^\circ\)
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