Question

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, $\pi \sqrt{3} \approx 5$ The value of $\phi$ is _____

Answer 1

Correct Answer: 60

Calculate the resistance of the lamp:
\(p=\frac{V^2}{R}\)
\(R=\frac{V^2}{p}=\frac{100^2}{500}\)
\(R=20\Omega\)
Calculate the current through the lamp:
\(I=\frac{V}{R}\)
\(I=\frac{100}{20}=5A\)

Impedance in the circuit:
Since the lamp and capacitor are in series, the impedance of the capacitor is given by:
\(X_C = \frac{1}{2 \pi f C}\)

Calculate the total impedance Z:
The total impedance Z of the series circuit is:
\(Z = \sqrt{R^2 + X_C^2}\)
But we need Z to find the phase angle. First, let's find X_C.
Calculate X_C:
We know the current I and the total voltage V_s:
\(V_s =IZ\)
\(Z=\frac{V_s}{I}=\frac{200}{5} ​=40Ω\)

Use Z to find X_C​:
\(Z^2 = R^2 + X_C^2\)
\(X_C​  =  \sqrt1200 ​\)

\(Given \ \pi \sqrt{3} \approx 5\)
\(X_C \approx \sqrt{400 \times 3} = \sqrt{400} \times \sqrt{3} = 20 \times \sqrt{3}\)
\(Using \sqrt{3} \approx 1.7323​\)
\(X_C \approx 20 \times 1.732 = 34.64 \, \Omega\)

Calculate the phase angle \(\phi\):
\(tanϕ=\frac{X_C​​}{R}=\frac{34.64}{20}≈1.732\)
\(We\ know\ that\ tan⁡60\degree = \sqrt{3} \approx 1.732\)
Hence
\(\phi \approx 60^\degree\)

So, the correct answer is \(60\degree\)