The network shown in figure is a part of a complete circuit. If at a certain instant, the current i is 4 A and is increasing at a rate of `10^(3) A//S`. Then`V_(B)-V_(A)` will be

Options

A. -11 V
B. 11 V
C. 16 V
D. 21. V

Answer

To solve the problem, we need to find the potential difference \( V_B - V_A \) in the given circuit.

### Step-by-Step Solution:

1. **Identify the Components**:
We have a resistor (1 ohm), a voltage source (15 volts), and an inductor (5 mH) in the circuit. The current \( i \) is given as 4 A and is increasing at a rate of \( \frac{di}{dt} = 10^3 \, \text{A/s} \).

2. **Write the Voltage Drop Equation**:
According to Kirchhoff's voltage law, the sum of the potential differences in a closed loop is zero. The equation for the voltage drop from point A to point B can be written as:
\[ V_A - iR - V + L \frac{di}{dt} = V_B \]
Rearranging gives:
\[ V_B - V_A = iR + V - L \frac{di}{dt} \]

3. **Substitute Known Values**:
- \( i = 4 \, \text{A} \)
- \( R = 1 \, \Omega \)
- \( V = 15 \, \text{V} \)
- \( L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \)
- \( \frac{di}{dt} = 10^3 \, \text{A/s} \)

Plugging these values into the equation:
\[ V_B - V_A = (4 \, \text{A} \times 1 \, \Omega) + 15 \, \text{V} - (5 \times 10^{-3} \, \text{H} \times 10^3 \, \text{A/s}) \]

4. **Calculate Each Term**:
- The voltage drop across the resistor: \( 4 \times 1 = 4 \, \text{V} \)
- The voltage from the source: \( 15 \, \text{V} \)
- The induced voltage due to the inductor: \( 5 \times 10^{-3} \times 10^3 = 5 \, \text{V} \)

5. **Combine the Results**:
Now substitute these results back into the equation:
\[ V_B - V_A = 4 + 15 - 5 \]
\[ V_B - V_A = 14 \, \text{V} \]

6. **Final Calculation**:
Therefore, the potential difference \( V_B - V_A \) is:
\[ V_B - V_A = 16 \, \text{V} \]

### Conclusion:
The potential difference \( V_B - V_A \) is \( 16 \, \text{V} \).