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 compute \(V_A - V_B\), we utilize the relationship between voltage, resistance, and inductance, incorporating Ohm's law and Faraday's law. For a circuit segment with resistance \(R\) and inductance \(L\) experiencing a changing current, the potential difference \(V\) is expressed as:

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

where:

• \(i\) represents the current in Amperes (A).
• \(\frac{di}{dt}\) is the rate of change of current in Amperes per second (A/s).
• \(R\) is the resistance in Ohms (\(\Omega\)).
• \(L\) is the inductance in Henries (H).

Given the parameters:

• \(R = 2 \, \Omega\)
• \(L = 4 \, \text{H}\)
• Current \(i = 2 \, \text{A}\)
• Rate of change of current \(\frac{di}{dt} = 1 \, \text{A/s}\)

Substituting these values into the formula yields:

\(V = (2 \, \text{A}) \cdot (2 \, \Omega) + (4 \, \text{H}) \cdot (1 \, \text{A/s})\)

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

Consequently, the calculated potential difference is:

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