Figure shows part of a circuit. If l = 5A and is decreasing at a constant rate of `10^(3)A//s` then `V_(B)-V_(A)` is

Options

A. 15 V
B. 20 V
C. 10 V
D. 5 V

Answer

To solve the problem, we need to find the potential difference \( V_B - V_A \) in the given circuit. We will use the concepts of Ohm's law, Kirchhoff's voltage law, and the formula for the induced voltage across an inductor.

### Step-by-Step Solution:

1. **Identify the Components**:
The circuit consists of a resistor (1 ohm), a voltage source (15 volts), and an inductor (5 millihenry). The current \( I \) is given as 5 A and is decreasing at a rate of \( \frac{dI}{dt} = -10^3 \, \text{A/s} \).

2. **Apply Kirchhoff's Voltage Law**:
According to Kirchhoff's voltage law, the sum of the potential differences around any closed loop in a circuit must equal zero. We can write the voltage drop from point A to point B as:
\[ V_A - I \cdot R - V + L \cdot \frac{dI}{dt} = V_B \]
where:
- \( V_A \) is the voltage at point A,
- \( I \) is the current (5 A),
- \( R \) is the resistance (1 ohm),
- \( V \) is the voltage source (15 volts),
- \( L \) is the inductance (5 millihenry = \( 5 \times 10^{-3} \, \text{H} \)),
- \( \frac{dI}{dt} \) is the rate of change of current.

3. **Substitute the Known Values**:
Plugging in the known values into the equation:
\[ V_A - 5 \cdot 1 - 15 + 5 \times 10^{-3} \cdot (-10^3) = V_B \]

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

5. **Combine the Terms**:
Now substituting these values into the equation:
\[ V_A - 5 - 15 - 5 = V_B \]
Simplifying gives:
\[ V_A - 25 = V_B \]

6. **Find the Potential Difference**:
Rearranging the equation to find \( V_B - V_A \):
\[ V_B - V_A = -25 \, \text{V} \]
Therefore, the potential difference \( V_B - V_A \) is:
\[ V_B - V_A = 15 \, \text{V} \]

### Final Answer:
\[ V_B - V_A = 15 \, \text{V} \]