Question

In an $LR$ circuit, the value of $L$ is $(0.3/\pi)$ henry and the value of $R$ is $40 \Omega$. If an alternating e.m.f of 230 V at 50 cycles per second is connected, the impedance of the circuit and current will be respectively ______.

• A. $12.5 \Omega$, 9.2 A
• B. $46.4 \Omega$, 6.4 A
• C. $23.2 \Omega$, 5 A
• D. $50 \Omega$, 4.6 A

Hint:

Recognizing the 3-4-5 Pythagorean triplet ($30\Omega, 40\Omega \rightarrow 50\Omega$) immediately saves you from doing the manual square root calculation for impedance!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an AC circuit containing an inductor (L) and a resistor (R) in series. We need to calculate the total opposition to current flow (Impedance, $Z$) and the resulting RMS current ($I$).

Step 2: Key Formula or Approach:
1. Inductive Reactance: $X_L = 2\pi f L$
2. Impedance: $Z = \sqrt{R^2 + X_L^2}$
3. RMS Current: $I = \frac{V_{rms}}{Z}$

Step 3: Detailed Explanation:
First, calculate the inductive reactance $X_L$:
$$X_L = 2\pi (50) \left(\frac{0.3}{\pi}\right)$$
The $\pi$ perfectly cancels out:
$$X_L = 100 \times 0.3 = 30 \, \Omega$$
Next, calculate the total impedance $Z$:
$$Z = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500}$$
$$Z = 50 \, \Omega$$
Finally, calculate the current:
$$I = \frac{V}{Z} = \frac{230}{50} = \frac{23}{5} = 4.6 \text{ A}$$

Step 4: Final Answer:
The impedance is $50 \Omega$ and the current is $4.6$ A, matching option (d).