Find the current in a circuit consisting of a coil and a capacitor in series with an A.C source of 110V (r.m.s.), 60Hz.The inductance of a coil in 0.80 H and its resistance is 50Ω.The capacitance of a capacitor is 8μF.

Answer

To find the current in a circuit consisting of a coil and a capacitor in series with an AC source, we will follow these steps:

### Step 1: Identify the given values
- Voltage (V_rms) = 110 V
- Frequency (f) = 60 Hz
- Inductance (L) = 0.80 H
- Resistance (R) = 50 Ω
- Capacitance (C) = 8 μF = 8 × 10^(-6) F

### Step 2: Calculate the capacitive reactance (X_C)
The formula for capacitive reactance is:
\[ X_C = \frac{1}{2 \pi f C} \]

Substituting the values:
\[ X_C = \frac{1}{2 \pi (60)(8 \times 10^{-6})} \]

Calculating:
\[ X_C = \frac{1}{2 \pi (60)(8 \times 10^{-6})} \approx 331.53 \, \Omega \]

### Step 3: Calculate the inductive reactance (X_L)
The formula for inductive reactance is:
\[ X_L = 2 \pi f L \]

Substituting the values:
\[ X_L = 2 \pi (60)(0.80) \]

Calculating:
\[ X_L = 2 \pi (60)(0.80) \approx 301.63 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The impedance in an RLC series circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Substituting the values:
\[ Z = \sqrt{(50)^2 + (301.63 - 331.53)^2} \]

Calculating:
1. Calculate \( X_L - X_C \):
\[ X_L - X_C = 301.63 - 331.53 = -29.90 \]
2. Now calculate \( Z \):
\[ Z = \sqrt{(50)^2 + (-29.90)^2} \]
\[ Z = \sqrt{2500 + 893.01} \]
\[ Z = \sqrt{3393.01} \approx 58.26 \, \Omega \]

### Step 5: Calculate the current (I_rms)
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]

Substituting the values:
\[ I_{rms} = \frac{110}{58.26} \]

Calculating:
\[ I_{rms} \approx 1.89 \, \text{A} \]

### Final Answer
The current in the circuit is approximately **1.89 A**.

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