A `100 Omega` resistance is connected in series with a `4 H` inductor. The voltage across the resistor is, `V_(R ) = (2.0 V) sin (10^(3) t)`.
Find amplitude of the voltage across the inductor.

Options

A. `40 V`
B. `60 V`
C. `80 V`
D. `90 V`

Explanation

To find the amplitude of the voltage across the inductor in a 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, \( V_R = 2.0 \, V \sin(10^3 t) \)

### Step 2: Determine the amplitude of the current
The voltage across the resistor can be expressed as:
\[ V_R = I \cdot R \]
Where \( I \) is the current in the circuit. The amplitude of the voltage across the resistor is given as \( 2.0 \, V \). Therefore, we can write:
\[ I_0 = \frac{V_{R0}}{R} \]
Substituting the values:
\[ I_0 = \frac{2.0 \, V}{100 \, \Omega} = 0.02 \, A = 2 \times 10^{-2} \, A \]

### Step 3: Determine the angular frequency
The angular frequency \( \omega \) can be extracted from the voltage equation:
\[ V_R = 2.0 \, V \sin(10^3 t) \]
Thus, \( \omega = 10^3 \, rad/s \).

### Step 4: Calculate the inductive reactance
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 10^3 \, rad/s \times 4 \, H = 4000 \, \Omega \]

### Step 5: Calculate the amplitude of the voltage across the inductor
The peak voltage across the inductor \( V_L \) can be calculated using Ohm's law for the inductor:
\[ V_L = I_0 \cdot X_L \]
Substituting the values:
\[ V_L = (2 \times 10^{-2} \, A) \cdot (4000 \, \Omega) = 80 \, V \]

### Conclusion
The amplitude of the voltage across the inductor is \( 80 \, V \).

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