Question

The magnitude of current $i$ in ampere unit is}

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $0.4$

Hint:

Use Kirchhoff's current and voltage laws to solve for currents in a circuit.

Answer 1

The Correct Option is A

To determine the magnitude of the current \( i \) in the given circuit, we need to analyze the arrangement of resistors and the flow of current in the circuit.

Let's break down the circuit:

• We have two parallel branches connected between points A and B.
• Branch 1 (Top): It includes resistors of \( 60 \, \Omega \) and parallel combination of two resistors \( 15 \, \Omega \) and \( 5 \, \Omega \).
• Branch 2 (Bottom): It includes a \( 10 \, \Omega \) resistor.

Given that \( 1 \, \text{A} \) current enters through both sides of the circuit, we can apply Kirchhoff's Current Law (KCL) which states that the sum of the currents entering a junction is equal to the sum of the currents leaving the junction.

Step-by-Step Solution:

1. Calculate the equivalent resistance of the resistors in the parallel combination \( 15 \, \Omega \) and \( 5 \, \Omega \):

\(R_{\text{parallel}} = \frac{1}{\frac{1}{15} + \frac{1}{5}} = \frac{1}{\frac{1 + 3}{15}} = \frac{15}{4} = 3.75 \, \Omega\)

1. Add this equivalent resistance to the \( 60 \, \Omega \) resistor to find the total resistance of Branch 1:

\(R_{\text{Branch 1}} = 60 + 3.75 = 63.75 \, \Omega\)

1. Since the total current entering the circuit is \( 2 \, \text{A} \) (from both sides), the current will split between the two branches inversely proportional to their resistances.
2. Use the formula for current division:

\(i_{\text{Branch 1}} = \frac{R_{\text{Branch 2}}}{R_{\text{Branch 1}} + R_{\text{Branch 2}}} \cdot I_{\text{total}}\)

\(i_{\text{Branch 1}} = \frac{10}{63.75 + 10} \cdot 2 \approx 0.1 \, \text{A}\)

Conclusion:

The magnitude of current \( i \) in the circuit is \(0.1 \, \text{A}\), which matches the correct option given.