An alternating emf of 220V is applied to a circuit containing a resistor R having resistance of 160…

An alternating emf of $220\text{V}$ is applied to a circuit containing a resistor R having resistance of $160\,\Omega$ and a capacitor ‘C’ in series. The current is found to lead the supply voltage by an angle $\theta = \tan^{-1}(3/4)$.

(i)[3]

Calculate: (1) The capacitive reactance (2) Impedance of the circuit (3) Current flowing in the circuit

(ii)[2]

If the frequency of the applied emf is $50\text{ Hz}$, what is the value of the capacitance of the capacitor ‘C’?

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Answer

Answer (i)

Formulas used: Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with R = 160 ohm, XC = 120 ohm, so Z = 200 ohm Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with V = 220 V, Z = 200 ohm, so I = 1.1 A

Given: $V = 220\text{ V}$, $R = 160\ \Omega$, $\tan\theta = \frac{3}{4}$. (1) $\tan\theta = \frac{X_C}{R}$, so $X_C = 160 \times \frac{3}{4} = 120\ \Omega$. (2) $Z = \sqrt{R^2 + X_C^2} = \sqrt{160^2 + 120^2} = 200\ \Omega$. (3) $I = \frac{V}{Z} = \frac{220}{200} = 1\cdot1\text{ A}$.

Final answer: $X_C = 120\ \Omega,\ Z = 200\ \Omega,\ I = 1.1\text{ A}$

Answer (ii)

Formula used: Capacitive reactance: XC = 1/(2*pi*f*C) with f = 50 Hz, XC = 120 ohm, so C = 2.65e-5 F

$X_C = \frac{1}{2\pi f C}$, so $C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 120} = 2\cdot65\times10^{-5}\text{ F} \approx 26\cdot5\ \mu\text{F}$.

Final answer: 2.65e-5 F