A `50 Omega ` resistor is connected in series with an inductor of 0.2 H and capacitor with capacitive reactance `40 Omega` . The combination is further connected to an AC source 180 V , 50 Hz. What will be the phase angle between current and voltage ? Also find the value of rms current in circuit.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 50 Ω
- Inductance (L) = 0.2 H
- Capacitive Reactance (Xc) = 40 Ω
- AC Voltage (Vrms) = 180 V
- Frequency (f) = 50 Hz

### Step 2: Calculate the Inductive Reactance (Xl)
Inductive reactance (Xl) can be calculated using the formula:
\[ X_l = 2 \pi f L \]

Substituting the values:
\[ X_l = 2 \pi (50) (0.2) \]
\[ X_l = 2 \pi (10) \]
\[ X_l = 62.83 \, \Omega \]

### Step 3: Calculate the net reactance (X)
Since the inductor and capacitor are in series, the net reactance (X) is given by:
\[ X = X_l - X_c \]
\[ X = 62.83 - 40 \]
\[ X = 22.83 \, \Omega \]

### Step 4: Calculate the Impedance (Z)
The total impedance (Z) of the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{(50)^2 + (22.83)^2} \]
\[ Z = \sqrt{2500 + 520.0889} \]
\[ Z = \sqrt{3020.0889} \]
\[ Z \approx 54.9 \, \Omega \]

### Step 5: Calculate the RMS current (Irms)
The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{180}{54.9} \]
\[ I_{rms} \approx 3.28 \, A \]

### Step 6: Calculate the phase angle (φ)
The phase angle (φ) between the current and voltage can be calculated using:
\[ \tan(\phi) = \frac{X}{R} \]
Substituting the values:
\[ \tan(\phi) = \frac{22.83}{50} \]
\[ \phi = \tan^{-1}\left(\frac{22.83}{50}\right) \]
Calculating φ:
\[ \phi \approx 24.6^\circ \]

### Final Answers
- The phase angle (φ) between current and voltage is approximately **24.6 degrees**.
- The RMS current (Irms) in the circuit is approximately **3.28 A**.