Question

A 0.07 H inductor and a 12 $\Omega$ resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as 22/7]

• A. 8.8 A and $\tan^{-1}(11/6)$
• B. 0.88 A and $\tan^{-1}(11/6)$
• C. 88 A and $\tan^{-1}(11/6)$
• D. 8.8 A and $\tan^{-1}(6/11)$

Hint:

For series AC circuits, use the impedance triangle. The base is resistance (R), the perpendicular is net reactance ($X = X_L - X_C$), and the hypotenuse is impedance (Z). The phase angle $\phi$ is given by $\tan(\phi) = X/R$.

Answer 1

The Correct Option is A

In a series LR circuit, we first calculate the inductive reactance ($X_L$).
$L = 0.07$ H, $f = 50$ Hz, $\pi = 22/7$.
$X_L = 2\pi fL = 2 \times \frac{22}{7} \times 50 \times 0.07 = 2 \times \frac{22}{7} \times 50 \times \frac{7}{100}$
$X_L = 2 \times 22 \times \frac{50}{100} = 44 \times \frac{1}{2} = 22 \, \Omega$
Next, we find the total impedance (Z) of the circuit.
$R = 12 \, \Omega$
$Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 22^2} = \sqrt{144 + 484} = \sqrt{628} \, \Omega$
Since $25^2 = 625$, we can approximate $Z \approx 25 \, \Omega$.
The RMS current ($I_{rms}$) is given by Ohm's law for AC circuits.
$V_{rms} = 220$ V.
$I_{rms} = \frac{V_{rms}}{Z} \approx \frac{220}{25} = \frac{220 \times 4}{100} = 8.8$ A.
The phase angle ($\phi$) between voltage and current is given by:
$\tan(\phi) = \frac{X_L}{R}$
$\tan(\phi) = \frac{22}{12} = \frac{11}{6}$
So, $\phi = \tan^{-1}\left(\frac{11}{6}\right)$. The voltage leads the current.
Thus, the current is approximately 8.8 A and the phase angle is $\tan^{-1}(11/6)$.