A `100 Omega` resistance is connected in series with a `4 H` inductor. The voltage across the resistor is `V_(R) = 2 sin (1000 t) V`. The voltage across the inductors is

Options

A. `80 sin (1000 t + (pi)/(2))`
B. `40 sin (1000 t + (pi)/(2))`
C. `80 sin (1000 t - (pi)/(2))`
D. `40 sin (1000 pi - (pi)/(2))`

Explanation

To find the voltage across the inductor in the given series circuit with a resistor and an inductor, we can follow these steps:

### Step 1: Identify the given values
- Resistance, \( R = 100 \, \Omega \)
- Inductance, \( L = 4 \, H \)
- Voltage across the resistor, \( V_R = 2 \sin(1000t) \, V \)

### Step 2: Calculate the peak voltage across the resistor
The voltage across the resistor can be expressed in terms of its peak value:
\[ V_{R_{peak}} = 2 \, V \]

### Step 3: Calculate the peak current in the circuit
Using Ohm's law, the peak current \( I_0 \) can be calculated as:
\[ I_0 = \frac{V_{R_{peak}}}{R} = \frac{2}{100} = 0.02 \, A \]

### Step 4: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = \omega L \]
where \( \omega = 1000 \, rad/s \) (from the voltage equation). Thus,
\[ X_L = 1000 \times 4 = 4000 \, \Omega \]

### Step 5: Calculate the peak voltage across the inductor \( V_{L_{peak}} \)
The peak voltage across the inductor can be calculated using:
\[ V_{L_{peak}} = I_0 \times X_L \]
Substituting the values we found:
\[ V_{L_{peak}} = 0.02 \times 4000 = 80 \, V \]

### Step 6: Determine the phase relationship
In a series \( R-L \) circuit, the voltage across the inductor leads the current by \( 90^\circ \). Therefore, the voltage across the inductor can be expressed as:
\[ V_L = V_{L_{peak}} \sin(\omega t + \frac{\pi}{2}) \]
Substituting the peak voltage:
\[ V_L = 80 \sin(1000t + \frac{\pi}{2}) \, V \]

### Final Answer
Thus, the voltage across the inductor is:
\[ V_L = 80 \sin(1000t + \frac{\pi}{2}) \, V \]