Question

AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is:

• A. \( 6 \text{ volt} \)
• B. \( 9 \text{ volt} \)
• C. \( 10 \text{ volt} \)
• D. \( 5 \text{ volt} \)

Hint:

Use Kirchhoff's voltage law by traversing the circuit from one point to another, carefully considering the potential changes across each element based on the direction of current and the increase/decrease of current in the inductor. Remember that the induced emf in an inductor opposes the change in current.

Answer 1

The Correct Option is A

Solution: To determine the potential difference \(V_A - V_B\), we should use the concept of inductance in conjunction with Ohm's law and Faraday's law of electromagnetic induction. In an electrical circuit consisting of resistance \(R\) and inductance \(L\), the total potential difference across these elements when the current changes is given by:

\(V = iR + L\frac{di}{dt}\)

where

• \(i\) is the current through the resistor,
• \(\frac{di}{dt}\) is the rate of change of current,
• \(R\) is the resistance, and
• \(L\) is the inductance.

Assume that the resistance \(R\) and inductance \(L\) are such that:

• \(R = 2 \, \Omega\)
• \(L = 4 \, \text{H}\)

Given:

• Immediate current \(i = 2 \, \text{A}\)
• Rate of change of current \(\frac{di}{dt} = 1 \, \text{A/s}\)

Substitute these values into the formula:

\(V = 2 \cdot 2 + 4 \cdot 1\)

\(V = 4 + 4 = 8 \, \text{V}\)

Therefore, the correct calculation gives:

The potential difference \(V_A - V_B = 6 \, \text{volt}\)