Question

A circuit containing inductance of $\frac{1}{6\pi}$H and a resistance of 15 $\Omega$ in series. If an AC voltage of 100 V and 60 Hz is applied to above circuit, then the current in the circuit and phase difference between voltage and current respectively are

• A. 4 A and $\tan^{-1}\left(\frac{4}{5}\right)$
• B. 5.3 A and $\tan^{-1}\left(\frac{3}{4}\right)$
• C. 4 A and $\tan^{-1}\left(\frac{4}{3}\right)$
• D. 5.3 A and $\tan^{-1}\left(\frac{4}{3}\right)$

Hint:

Inductive reactance: $X_L = 2\pi fL$.
Impedance of series LR circuit: $Z = \sqrt{R^2 + X_L^2}$.
RMS current: $I_{rms} = V_{rms}/Z$. (Assume given AC voltage is RMS unless specified otherwise).
Phase angle for series LR circuit: $\tan\phi = X_L/R$. Voltage leads current.
Look for Pythagorean triples (e.g., 3-4-5, 5-12-13) to simplify $\sqrt{R^2+X_L^2}$. Here $15, 20 \rightarrow 25$ is $5 \times (3,4,5)$.

Answer 1

The Correct Option is C

This is a series LR circuit connected to an AC voltage source. Given values: Inductance $L = \frac{1}{6\pi}$ H. Resistance $R = 15 \text{ } \Omega$. AC voltage $V_{rms} = 100 \text{ V}$ (assuming RMS, as is standard for AC voltage values). Frequency $f = 60 \text{ Hz}$. 1. Calculate Inductive Reactance ($X_L$): $X_L = 2\pi fL$. $X_L = 2\pi \times (60 \text{ Hz}) \times \left(\frac{1}{6\pi} \text{ H}\right)$. $X_L = 2\pi \times 60 \times \frac{1}{6\pi} = \frac{120\pi}{6\pi} = 20 \text{ } \Omega$. 2. Calculate Impedance ($Z$): For a series LR circuit, $Z = \sqrt{R^2 + X_L^2}$. $Z = \sqrt{(15 \text{ } \Omega)^2 + (20 \text{ } \Omega)^2}$. $Z = \sqrt{225 + 400} = \sqrt{625}$. $Z = 25 \text{ } \Omega$. (This is a 3-4-5 Pythagorean triple scaled by 5: $15=3\times 5$, $20=4\times 5$, so $Z=5\times 5=25$). 3. Calculate Current ($I_{rms}$): $I_{rms} = \frac{V_{rms}}{Z}$. $I_{rms} = \frac{100 \text{ V}}{25 \text{ } \Omega} = 4 \text{ A}$. 4. Calculate Phase Difference ($\phi$): The phase difference $\phi$ between voltage and current in a series LR circuit is given by $\tan\phi = \frac{X_L}{R}$. In an LR circuit, voltage leads current (or current lags voltage). $\tan\phi = \frac{20 \text{ } \Omega}{15 \text{ } \Omega} = \frac{20}{15} = \frac{4}{3}$. So, $\phi = \tan^{-1}\left(\frac{4}{3}\right)$. The current in the circuit is 4 A and the phase difference is $\tan^{-1}\left(\frac{4}{3}\right)$. This matches option (c). \[ \boxed{\text{4 A and } \tan^{-1}\left(\frac{4}{3}\right)} \]