Let f = 50 Hz, and C = 100 `muF` in an AC circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A at t = 0. The expression for the instantaneous voltage across the capacitor will be

Options

A. `C=50sin(100pit-pi//2)`
B. `C=100 sin(50pit)`
C. `C=50sin100pit`
D. `C=50sin(100pit+pi//2)`

Answer

To find the expression for the instantaneous voltage across the capacitor in the given AC circuit, we can follow these steps:

### Step 1: Identify the given values
- Frequency (f) = 50 Hz
- Capacitance (C) = 100 µF = 100 × 10^(-6) F
- Peak current (I₀) = 1.57 A

### Step 2: Calculate angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \times 50 = 100\pi \text{ rad/s} \]

### Step 3: Calculate capacitive reactance (Xₐ)
The capacitive reactance (Xₐ) is given by:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values of ω and C:
\[ X_C = \frac{1}{100\pi \times 100 \times 10^{-6}} = \frac{1}{0.01\pi} = \frac{100}{\pi} \text{ ohms} \]

### Step 4: Calculate peak voltage (V₀)
Using the relationship between peak current (I₀), peak voltage (V₀), and capacitive reactance (Xₐ):
\[ I₀ = \frac{V₀}{X_C} \]
Rearranging gives:
\[ V₀ = I₀ \times X_C \]
Substituting the values:
\[ V₀ = 1.57 \times \frac{100}{\pi} \]
Calculating this gives:
\[ V₀ \approx \frac{157}{\pi} \approx 50 \text{ volts} \]

### Step 5: Write the expression for instantaneous voltage
In a purely capacitive circuit, the voltage lags the current by 90 degrees (or π/2 radians). Therefore, the instantaneous voltage (V(t)) can be expressed as:
\[ V(t) = V₀ \sin(\omega t - \frac{\pi}{2}) \]
Substituting the values of V₀ and ω:
\[ V(t) = 50 \sin(100\pi t - \frac{\pi}{2}) \]

### Final Expression
Thus, the expression for the instantaneous voltage across the capacitor is:
\[ V(t) = 50 \sin(100\pi t - \frac{\pi}{2}) \]

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