Question

An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ______ × 10² Hz

Answer 1

Correct Answer: 5

To find the frequency of the AC source when the current is in phase with the EMF, we must have the total impedance of the circuit purely resistive. This occurs when the inductive reactance X_L equals the capacitive reactance X_C. 

X_L=X_C implies 2πfL = 1/(2πfC), where: L = 0.5 \text{ mH} = 0.5 \times 10^{-3} \text{ H} and C = 200 \text{ μF} = 200 \times 10^{-6} \text{ F}.

Setting the equations:

2πf(0.5 \times 10^{-3}) = 1/(2πf \times 200 \times 10^{-6})

Solving for f:

(2πf)^2 = 1/(0.5 \times 10^{-3})(200 \times 10^{-6})

Simplifying gives:

4π^2f^2 = 1/(0.5 \times 10^{-3} \times 200 \times 10^{-6})

Calculate the right-hand side:

1/(0.5 \times 200 \times 10^{-9}) = 1/(100 \times 10^{-9}) = 10^{7}

Thus,

4π^2f^2 = 10^{7}

Solving for f^2:

f^2 = 10^{7}/4π^2

f = \sqrt{10^{7}/4π^2}

Calculate f:

f ≈ √(10^{7}/39.4784) ≈ 503.3

This gives f ≈ 5 \times 10^{2} \text{ Hz}

The calculated frequency f is approximately 5 \times 10^{2} \text{ Hz}, which lies within the specified range of 5,5 × 10².