Two loops share the resistor R_2 = 3 ohms. Loop 1 contains battery E_1 = 18 V and resistor R_1 = 6 ohms; loop 2 contains battery E_2 = 12 V and resistor R_3 = 6 ohms. Both mesh currents I_1 and I_2 are taken clockwise, so the current through R_2 is I_1 - I_2. By Kirchhoff's laws, what is the current through R_2?

Answer and explanation

Correct answer: B. 0.5 A

  1. Step 1: Apply Kirchhoff's voltage law to the two meshes.
  2. Step 2: Loop 1: 18 - 6I_1 - 3(I_1 - I_2) = 0, which gives 9I_1 - 3I_2 = 18, or 3*I_1 - I_2 = 6.
  3. Step 3: Loop 2: 12 - 6I_2 - 3(I_2 - I_1) = 0, which gives -3I_1 + 9I_2 = 12, or -I_1 + 3*I_2 = 4.
  4. Step 4: Solve the system: from loop 1, I_2 = 3I_1 - 6. Substitute into loop 2: -I_1 + 3(3I_1 - 6) = 4, yielding 8I_1 = 22, so I_1 = 2.75 A.
  5. Step 5: Then I_2 = 3*(2.75) - 6 = 2.25 A.
  6. Step 6: Current through middle resistor R_2: I = I_1 - I_2 = 2.75 - 2.25 = 0.5 A.
Answer
0.5 A.
Why not the others: 1.0 A assumes symmetric branch currents without cross-mesh coupling; 1.5 A calculates using only battery 1; 2.0 A neglects the opposing mesh potential.

Practise more Electromagnetism, Wave Optics and Modern Physics.

Free questions with an explanation for every one, and full timed Bagrut (Secondary School Matriculation) mock tests.

All Electromagnetism, Wave Optics and Modern Physics questionsTake a mock test

More Electromagnetism, Wave Optics and Modern Physics questions

Revise with flashcards