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 calculate the current through the \( 20 \, \Omega \) resistor, we proceed as follows:

 Step 1: Identify currents and apply Kirchhoff's Current Law (KCL)
Define the currents through the \( 10 \, \Omega \), \( 20 \, \Omega \), and \( 30 \, \Omega \) resistors as \( I_1, I_2, \text{and} \, I_3 \), respectively. At the junction point, KCL states: \[ I_1 = I_2 + I_3. \] Using Ohm's Law, the currents are expressed as: \[ I_1 = \frac{V_a - V_b}{10}, \quad I_2 = \frac{V_b - V_c}{20}, \quad I_3 = \frac{V_b - V_c}{30}. \] 

Step 2: Substitute known voltage values
The given potentials are: \[ V_a = 30 \, \text{V}, \quad V_b = 12 \, \text{V}, \quad V_c = 2 \, \text{V}. \] Substituting these into the 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: Verify current conservation at the junction
Check if \( I_1 = I_2 + I_3 \): \[ I_1 = 0.4 + 0.333 = 1.8 \, \text{A}. \] This confirms that Kirchhoff's Current Law is satisfied. 

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

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