NEB Solutions: Alternating Currents

[2076 ‘C’]

A circuit consists of a capacitor of 2 μF and a resistor of 1000 Ω. An alternating emf of 12 V(rms) and frequency 50 Hz is applied. Find the current flowing, the voltage across capacitor and the phase angel between the applied emf and current.

Solution
Given,
Capacitance (C) = 2μF = [latex]2 \times 10^{-6}[/latex] F
Resistance (R) = 1000 Ω
Emf (E) = 12 V
Frequency (f) = 50 Hz
Current flowing (I) = ?
Voltage across capacitor [latex](V_c)[/latex] = ?
Phase angle (ϕ) = ?

We know that,
Capacitive reactance [latex](X_c) = \frac{1}{2\pi f C} = \frac{1}{2\times 3.14\times 50\times (2\times 10^{-6})} = 1592.357\ \Omega[/latex]

Total impedance [latex](Z) = \sqrt{R^2+X_C^2} = \sqrt{1000^2+1592.357^2} = 1880.319\ \Omega[/latex]

Now, total current flowing in the circuit [latex](I) = \frac{E}{Z} = \frac{12}{1880.319} = 6.382\times 10^{-3}\ A[/latex]

Again, voltage across capacitor [latex](V_c) = IX_C = (6.382\times 10^{-3})\times 1592.357 = 10.162\ V[/latex]

Now, phase angle [latex](tan\phi) = \frac{V_C}{V_R} = \frac{IX_C}{IR} = \frac{X_C}{R} = \frac{1592.357}{1000} = 1.592[/latex]

Or, [latex]\phi = \tan^{-1}(1.592) = 57.865^{\circ}[/latex].

[2075 ‘S’]

An ac source of 220 V, 50 Hz is connected to series circuit containing a resistor R an inductor L and a capacitor C. If R = 200 Ω, L = 0.5 H and C = 10 μF, calculate (i) the current in the circuit (ii) the phase angle and (iii) the power consumed in the circuit.

Solution
Given,
Emf (E) = 220 V
Frequency (f) = 50 Hz
R = 200 Ω, L = 0.5 H and C = 10 μF = [latex]10\times 10^{-6}F[/latex]
Current in the circuit (I) = ?
Phase angle (ϕ) = ?
Power consumed (P) = ?

We have,
Inductive reactance [latex](X_L) = 2\pi fL = 2\times 3.14\times 50\times 0.5 = 157\ \Omega[/latex]

Capacitive reactance [latex](X_C) = \frac{1}{2\pi f C} = \frac{1}{2\times 3.14\times 50\times (10\times 10^{-6})} = 318.471\ \Omega[/latex]