A `100 mu F` capacitor in series with a `40 Omega` resistor is connected to a 100 V, 60 Hz supply. The maximum current in the circuit is

Options

A. `2.65 A`
B. `2.75 A`
C. `2.85 A`
D. `2.95 A`

Explanation

To find the maximum current in the circuit with a capacitor and resistor in series connected to an AC supply, we will follow these steps:

### Step 1: Identify the given values
- Capacitance, \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- Resistance, \( R = 40 \, \Omega \)
- Voltage (RMS), \( V_{rms} = 100 \, V \)
- Frequency, \( f = 60 \, Hz \)

### Step 2: Calculate the capacitive reactance (\( X_c \))
The formula for capacitive reactance is given by:
\[ X_c = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_c = \frac{1}{2 \pi \times 60 \times 100 \times 10^{-6}} \]
Calculating this gives:
\[ X_c \approx 26.53 \, \Omega \]

### Step 3: Calculate the impedance (\( Z \))
The impedance in a series circuit with a resistor and a capacitor is given by:
\[ Z = \sqrt{R^2 + X_c^2} \]
Substituting the values:
\[ Z = \sqrt{40^2 + 26.53^2} \]
Calculating this gives:
\[ Z \approx \sqrt{1600 + 703.51} \approx \sqrt{2303.51} \approx 47.99 \, \Omega \]

### Step 4: Calculate the peak voltage (\( V_0 \))
The peak voltage can be calculated from the RMS voltage using the relation:
\[ V_0 = V_{rms} \times \sqrt{2} \]
Substituting the values:
\[ V_0 = 100 \times \sqrt{2} \approx 100 \times 1.414 \approx 141.42 \, V \]

### Step 5: Calculate the maximum current (\( I_0 \))
The maximum current can be calculated using Ohm's law for AC circuits:
\[ I_0 = \frac{V_0}{Z} \]
Substituting the values:
\[ I_0 = \frac{141.42}{47.99} \approx 2.94 \, A \]

### Final Answer
The maximum current in the circuit is approximately:
\[ I_0 \approx 2.94 \, A \]