Question

(a) A series circuit having resistance, Inductance and capacitance is to be operated on a constant voltage supply of available frequency. Indicate graphically how changes will take place in the resistive terms, reactive terms, i.e. capacitive reactance and inductive reactance. (6)

(b) A resistance of 130 Ω and a capacitor of 30µF are connected in parallel across a 230 Volt, 50Hz supply. Find the current in each component, total current, phase angle and the power consumed. (10)

Answer 1

Part (a)

Series RLC circuit on constant voltage supply of variable frequency:

The resistance R is independent of frequency; it is a horizontal straight line on the graph of reactance/resistance against frequency.
Inductive reactance XL = 2 pi f L increases linearly with frequency (a straight line through the origin, rising with f).
Capacitive reactance XC = 1/(2 pi f C) decreases with frequency (a hyperbola, falling as f increases).
The net reactance X = XL - XC. At low frequency XC dominates (capacitive); at high frequency XL dominates (inductive).
At the resonant frequency f0, XL = XC, the net reactance is zero, and the circuit behaves as purely resistive. The impedance is a minimum (equal to R) and the current is a maximum.
Below resonance the circuit is capacitive (current leads voltage); above resonance it is inductive (current lags voltage).
The graph shows R as a horizontal line, XL rising linearly, XC falling hyperbolically, and the two curves crossing at resonance.

Part (b)

Resistance 130 ohm and capacitor 30 uF in parallel across 230 V, 50 Hz:

Capacitive reactance XC = 1/(2 pi f C) = 1/(2 x 3.1416 x 50 x 30 x 10^-6) = 1/0.009425 = 106.1 ohm.
Current in resistor: IR = V/R = 230/130 = 1.769 A (in phase with voltage).
Current in capacitor: IC = V/XC = 230/106.1 = 2.168 A (leads voltage by 90 degrees).
Total current: I = sqrt(IR^2 + IC^2) = sqrt(1.769^2 + 2.168^2) = sqrt(3.129 + 4.700) = sqrt(7.829) = 2.798 A.
Phase angle: cos phi = IR/I = 1.769/2.798 = 0.632, so phi = 50.8 degrees (current leading voltage, since capacitive).
Power consumed: P = V x IR = 230 x 1.769 = 406.9 W (only the resistor consumes power; also P = V I cos phi = 230 x 2.798 x 0.632 = 406.9 W).

So IR = 1.77 A, IC = 2.17 A, total current = 2.80 A, phase angle = 50.8 degrees leading, power = 407 W.

✓ Verified Model Answer (Text Solution)

Series RLC circuit on constant voltage supply of variable frequency:

Resistance 130 ohm and capacitor 30 uF in parallel across 230 V, 50 Hz: