A coil of resistance 10 Ω and inductance 47.7 mH is connected to a 200 V, 50 Hz supply. Calculate (a) the current drawn from the supply, (b) the power factor of the circuit. 5 A capacitor of 50 µF capacitance is connected in series with a resistance of 50 Ω to a 200 V, 50 Hz supply. Calculate (a) the current drawn from the supply, (b) the voltage across the resistance, and (c) the voltage across the capacitance.
The Correct Answer and Explanation is :
Solution to Problem 1:
A coil with resistance (R) of 10 Ω and inductance (L) of 47.7 mH is connected to a 200 V, 50 Hz supply.
(a) Current drawn from the supply:
- Calculate inductive reactance ((X_L)):
[
X_L = 2 \pi f L
]
Substituting (f = 50 \, \text{Hz}) and (L = 47.7 \, \text{mH} = 0.0477 \, \text{H}):
[
X_L = 2 \pi (50)(0.0477) = 14.96 \, \Omega
] - Calculate total impedance ((Z)):
[
Z = \sqrt{R^2 + X_L^2}
]
Substituting (R = 10 \, \Omega) and (X_L = 14.96 \, \Omega):
[
Z = \sqrt{10^2 + 14.96^2} = \sqrt{100 + 223.81} = \sqrt{323.81} \approx 18.0 \, \Omega
] - Calculate current ((I)):
[
I = \frac{V}{Z}
]
Substituting (V = 200 \, \text{V}) and (Z = 18.0 \, \Omega):
[
I = \frac{200}{18.0} \approx 11.11 \, \text{A}
]
(b) Power factor (( \text{pf} )):
- Calculate power factor:
[
\text{pf} = \frac{R}{Z}
]
Substituting (R = 10 \, \Omega) and (Z = 18.0 \, \Omega):
[
\text{pf} = \frac{10}{18.0} \approx 0.556
]
Solution to Problem 2:
A capacitor of 50 µF and a resistance of 50 Ω are connected to a 200 V, 50 Hz supply.
(a) Current drawn from the supply:
- Calculate capacitive reactance ((X_C)):
[
X_C = \frac{1}{2 \pi f C}
]
Substituting (f = 50 \, \text{Hz}) and (C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F}):
[
X_C = \frac{1}{2 \pi (50)(50 \times 10^{-6})} \approx 63.66 \, \Omega
] - Calculate total impedance ((Z)):
[
Z = \sqrt{R^2 + X_C^2}
]
Substituting (R = 50 \, \Omega) and (X_C = 63.66 \, \Omega):
[
Z = \sqrt{50^2 + 63.66^2} = \sqrt{2500 + 4053.63} \approx \sqrt{6553.63} \approx 80.91 \, \Omega
] - Calculate current ((I)):
[
I = \frac{V}{Z}
]
Substituting (V = 200 \, \text{V}) and (Z = 80.91 \, \Omega):
[
I = \frac{200}{80.91} \approx 2.47 \, \text{A}
]
(b) Voltage across the resistance:
[
V_R = I R
]
Substituting (I = 2.47 \, \text{A}) and (R = 50 \, \Omega):
[
V_R = 2.47 \times 50 \approx 123.5 \, \text{V}
]
(c) Voltage across the capacitance:
[
V_C = I X_C
]
Substituting (I = 2.47 \, \text{A}) and (X_C = 63.66 \, \Omega):
[
V_C = 2.47 \times 63.66 \approx 157.3 \, \text{V}
]
Explanation:
In both problems, we used the concept of impedance in AC circuits, which combines resistance and reactance (inductive or capacitive). In Problem 1, the inductive coil’s impedance is dominated by its inductive reactance, causing a lagging power factor. In Problem 2, the capacitive circuit’s impedance depends on the balance between resistance and capacitive reactance, where the voltage across components is calculated using Ohm’s Law and the reactance formula.