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 and phase angle of this current
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 and phase angle of this current
Solution
To solve this problem, we need to use the formulas for impedance and current in an AC circuit.
Step 1: Calculate the Inductive Reactance (XL) The formula for inductive reactance is XL = 2πfL, where f is the frequency and L is the inductance.
XL = 2π * 50Hz * 140mH = 2π * 50 * 0.14 = 44Ω
Step 2: Calculate the Impedance (Z) of the coil The impedance 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) = √(100 + 1936) = √2036 = 45.1Ω
Step 3: Calculate the Current in the Coil (I) The current in the coil can be found using Ohm's law (I = V/Z), where V is the voltage and Z is the impedance.
I = 230V / 45.1Ω = 5.1A
Step 4: Calculate the Phase Angle (φ) The phase angle can be found using the formula φ = arctan(XL/R), where XL is the reactance and R is the resistance.
φ = arctan(44Ω / 10Ω) = 77.47°
So, the current in the coil is 5.1A and the phase angle of this current is 77.47°.
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