Question

Find current and voltage across each resistor.

Answer 1

• Left Branch:

  • Current: 2 A through each resistor.
  • Voltage across each 3 Ω resistor: 6 V.

• Right Branch:

  • Total branch current: 4/3 A.
  • Voltage across the 6 Ω resistor: 8 V.
  • In the parallel combination:
    • Voltage across both resistors: 4 V.
    • Current through 12 Ω resistor: 1/3 A.
    • Current through 4 Ω resistor: 1 A.

Answer 2

1. Left Branch (Series 3 Ω and 3 Ω):

   • Total resistance:

     $$R_A = 3 + 3 = 6\,\Omega$$

   • Current in branch:

     $$I_A = \frac{V}{R_A} = \frac{12}{6} = 2\,\text{A}$$

   • Voltage drop across each resistor:

     $$V_{3\Omega} = 2\,\text{A} \times 3\,\Omega = 6\,\text{V}$$

2. Right Branch:

   • The branch consists of a 6 Ω resistor in series with a parallel combination of 12 Ω and 4 Ω.

   • First, find the equivalent resistance of the parallel resistors:

     $$R_{parallel} = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3\,\Omega$$

   • Total resistance in right branch:

     $$R_B = 6 + 3 = 9\,\Omega$$

   • Current in branch:

     $$I_B = \frac{12}{9} = \frac{4}{3}\,\text{A} \approx 1.33\,\text{A}$$

   • Voltage drop across 6 Ω resistor:

     $$V_{6\Omega} = \frac{4}{3}\,\text{A} \times 6\,\Omega = 8\,\text{V}$$

   • Voltage across the parallel combination:

     $$V_{parallel} = 12\,\text{V} - 8\,\text{V} = 4\,\text{V}$$

   • Currents in the parallel resistors:

     • Through 12 Ω:

       $$I_{12\Omega} = \frac{4}{12} = \frac{1}{3}\,\text{A} \approx 0.33\,\text{A}$$

     • Through 4 Ω:

       $$I_{4\Omega} = \frac{4}{4} = 1\,\text{A}$$

   • Verification:

     $$I_{12\Omega} + I_{4\Omega} = \frac{1}{3} + 1 = \frac{4}{3}\,\text{A} = I_B$$