A coil of resistance 10Ω and inductance 140mH is connected in parallel with a 260Ωresistor across a 230V, 50Hz supply. Calculate the following(i) Current in the coil (3 Marks)(ii) Supply current (3 Marks)(iii) Circuit impedance (3 Marks)(iv) Power factor (3 Marks)(v) Power consumed
Question
A coil of resistance 10Ω and inductance 140mH is connected in parallel with a 260Ωresistor across a 230V, 50Hz supply. Calculate the following(i) Current in the coil (3 Marks)(ii) Supply current (3 Marks)(iii) Circuit impedance (3 Marks)(iv) Power factor (3 Marks)(v) Power consumed
Solution
(i) Current in the coil:
First, we need to calculate the inductive reactance (XL) of the coil. The formula for inductive reactance is XL = 2πfL, where f is the frequency and L is the inductance.
XL = 2π * 50Hz * 140mH = 44Ω
The total impedance (Z) of the coil is the square root of the sum of the resistance squared and the reactance squared (Z = √(R² + XL²)).
Z = √((10Ω)² + (44Ω)²) = 45.8Ω
The current in the coil (I) is the voltage divided by the impedance (I = V/Z).
I = 230V / 45.8Ω = 5.02A
(ii) Supply current:
The supply current is the sum of the current in the coil and the current in the resistor. The current in the resistor (Ir) is the voltage divided by the resistance (Ir = V/R).
Ir = 230V / 260Ω = 0.88A
The supply current (Is) is the sum of I and Ir.
Is = 5.02A + 0.88A = 5.9A
(iii) Circuit impedance:
The circuit impedance (Zt) is the total voltage divided by the total current (Zt = V/Is).
Zt = 230V / 5.9A = 39Ω
(iv) Power factor:
The power factor (PF) is the resistance of the coil divided by the impedance of the coil (PF = R/Z).
PF = 10Ω / 45.8Ω = 0.22
(v) Power consumed:
The power consumed (P) is the voltage times the total current times the power factor (P = V * Is * PF).
P = 230V * 5.9A * 0.22 = 299.58W
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