A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage `V=10 sin (100t)`. The peak current in the circuit will be:

Options

A. `2` amp
B. `1` amp
C. `10` amp
D. `20` amp

Explanation

To find the peak current in the circuit with the given parameters, we can follow these steps:

### Step 1: Identify the given values
- Inductance \( L = 5.0 \, \text{mH} = 5.0 \times 10^{-3} \, \text{H} \)
- The voltage is given as \( V(t) = 10 \sin(100t) \)
- The peak voltage \( V_0 = 10 \, \text{V} \)
- The angular frequency \( \omega = 100 \, \text{rad/s} \)

### Step 2: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \times (5.0 \times 10^{-3}) = 0.5 \, \Omega \]

### Step 3: Calculate the peak current \( I_P \)
The peak current \( I_P \) can be calculated using Ohm's law for AC circuits:
\[ I_P = \frac{V_0}{X_L} \]
Substituting the peak voltage and the inductive reactance:
\[ I_P = \frac{10}{0.5} = 20 \, \text{A} \]

### Final Answer
The peak current in the circuit is \( 20 \, \text{A} \).

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