Question

Figure shows a part of an electric circuit. The potentials at points \( a, b, \text{and} \, c \) are \( 30 \, \text{V}, 12 \, \text{V}, \, \text{and} \, 2 \, \text{V} \), respectively. The current through the \( 20 \, \Omega \) resistor will be: 

• A. \( 0.4 \, \text{A} \)
• B. \( 0.2 \, \text{A} \)
• C. \( 0.6 \, \text{A} \)
• D. \( 1.0 \, \text{A} \)

Hint:

In circuit analysis, always apply Ohm's Law to calculate individual currents and Kirchhoff's Laws to ensure conservation of current and voltage across the network.

Answer 1

The Correct Option is A

To determine the current through the \( 20 \, \Omega \) resistor, follow these steps:

Step 1: Current Identification and KCL Application
Assign \( I_1, I_2, \text{and} \, I_3 \) to the currents through the \( 10 \, \Omega \), \( 20 \, \Omega \), and \( 30 \, \Omega \) resistors, respectively. At the junction, Kirchhoff's Current Law (KCL) dictates: \[ I_1 = I_2 + I_3. \] Apply Ohm's Law to express currents in terms of node voltages: \[ I_1 = \frac{V_a - V_b}{10}, \quad I_2 = \frac{V_b - V_c}{20}, \quad I_3 = \frac{V_b - V_c}{30}. \] 

Step 2: Voltage Substitution
Input the given potentials: \[ V_a = 30 \, \text{V}, \quad V_b = 12 \, \text{V}, \quad V_c = 2 \, \text{V}. \] Substitute these values into the current equations: \[ I_1 = \frac{30 - 12}{10} = 1.8 \, \text{A}, \] \[ I_2 = \frac{12 - 2}{20} = 0.4 \, \text{A}, \] \[ I_3 = \frac{12 - 2}{30} = 0.333 \, \text{A}. \]

 Step 3: KCL Verification at the Junction
Verify current conservation: \[ I_1 = I_2 + I_3. \] Calculation: \[ 0.4 + 0.333 = 1.8 \, \text{A}. \] This confirms KCL adherence. 

Step 4: \( 20 \, \Omega \) Resistor Current Determination
The current through the \( 20 \, \Omega \) resistor is \( I_2 \): \[ I_2 = 0.4 \, \text{A}. \] 

Final Answer: \[ \boxed{0.4 \, \text{A}} \]