A resistor of `12 Omega`, a capacitor of reactance `14 Omega` and a pure inductor of inductance 0.1 Hz are joined in series and placed across a 200 V, 50 Hz a.c. supply. Calculate (i) The current in the circuit and (ii) The phase angle between the current and the voltage. Take `pi = 3`.

Answer

To solve the problem step by step, we will first calculate the inductive reactance, then the impedance of the circuit, followed by the current and the phase angle.

### Step 1: Calculate the Inductive Reactance (XL)

The inductive reactance (XL) is given by the formula:
\[ XL = \omega L \]
where
\(\omega = 2\pi f\) and \(f\) is the frequency.

Given:
- \(L = 0.1 \, \text{H}\)
- \(f = 50 \, \text{Hz}\)
- \(\pi = 3\)

Calculating \(\omega\):
\[ \omega = 2 \times 3 \times 50 = 300 \, \text{rad/s} \]

Now, substituting \(\omega\) into the formula for \(XL\):
\[ XL = 300 \times 0.1 = 30 \, \Omega \]

### Step 2: Calculate the Impedance (Z)

The impedance (Z) of a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (XL - XC)^2} \]

Given:
- \(R = 12 \, \Omega\)
- \(XC = 14 \, \Omega\)
- \(XL = 30 \, \Omega\)

Now substituting the values:
\[ Z = \sqrt{12^2 + (30 - 14)^2} \]
Calculating:
\[ Z = \sqrt{144 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \Omega \]

### Step 3: Calculate the Current (I)

The current (I) in the circuit can be calculated using Ohm's law:
\[ I = \frac{V}{Z} \]

Given:
- \(V = 200 \, \text{V}\)
- \(Z = 20 \, \Omega\)

Now substituting the values:
\[ I = \frac{200}{20} = 10 \, \text{A} \]

### Step 4: Calculate the Phase Angle (φ)

The phase angle (φ) between the current and voltage is given by:
\[ \tan(\phi) = \frac{XL - XC}{R} \]

Substituting the known values:
\[ \tan(\phi) = \frac{30 - 14}{12} = \frac{16}{12} = \frac{4}{3} \]

Now, we can find φ using the arctangent function:
\[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \]

Using the known value:
\[ \phi \approx 53.13^\circ \]

### Final Answers:
1. The current in the circuit is \(10 \, \text{A}\).
2. The phase angle between the current and the voltage is \(53.13^\circ\).

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