Question

The network shown in figure is a part of complete circuit. What is a potential difference $V_A - V_B$ when the current I is 5A and is decreasing at a rate of $10^3 A/s$?

Answer 1

30V

Answer 2

To find the potential difference $V_A - V_B$, we can apply Kirchhoff's Voltage Law (KVL) by traversing the circuit from point A to point B.

The components in the circuit are:

1. A resistor with resistance $R = 1 \Omega$.
2. A voltage source (battery) with EMF $V_{batt} = 30 V$.
3. An inductor with inductance $L = 5 mH = 5 \times 10^{-3} H$.

The given current is $I = 5 A$. The rate of change of current is $\frac{dI}{dt} = -10^3 A/s$ (since the current is decreasing).

Let's apply KVL starting from $V_A$ and moving towards $V_B$:

1. Across the resistor ($R=1\Omega$): The current $I$ flows from A towards B. When current flows through a resistor, there is a potential drop in the direction of current. Potential change = $-I \times R = - (5 A) \times (1 \Omega) = -5 V$.

2. Across the voltage source ($30V$): The current $I$ flows from left to right. Observing the battery's polarity, the current enters the positive terminal and leaves the negative terminal. This indicates a potential drop across the battery (as if the battery is being charged or opposing the current flow). Potential change = $-30 V$.

3. Across the inductor ($L=5mH$): The voltage across an inductor is given by $V_L = L \frac{dI}{dt}$. When moving in the direction of current, the potential change is $-L \frac{dI}{dt}$. Potential change = $-L \frac{dI}{dt} = - (5 \times 10^{-3} H) \times (-10^3 A/s)$. Potential change = $- (-5 V) = +5 V$. (The positive sign indicates a potential gain across the inductor because the current is decreasing, and the inductor induces an EMF to oppose this decrease, effectively acting as a source in the direction of current flow.)

Now, sum all the potential changes from A to B: $V_A + (\text{change across R}) + (\text{change across battery}) + (\text{change across L}) = V_B$ $V_A - 5 V - 30 V + 5 V = V_B$ $V_A - 30 V = V_B$ Rearranging the terms to find $V_A - V_B$: $V_A - V_B = 30 V$