The value of current in the 6 Ω resistance is :

Solution

Physics sir · 4 steps

Ground the bottom wire and find the one unknown, the voltage at the middle junction; the 6 Ω current follows.

1. Set up potentials. Take the bottom wire as 0 V. Both batteries have their positive plate on top, so the top-left corner is at 140 V and the top-right corner is at 90 V. Let the top middle junction be at potential V.
2. Junction rule. Write the current leaving the junction through each of its three resistors. These currents must add to zero.

   (V-140)/20 + V/6 + (V-90)/5 = 0

3. Solve for V. Multiply through by 60: 3V - 420 + 10V + 12V - 1080 = 0, so 25V = 1500 and V = 60 V.
4. So. The current in the 6 Ω resistor is 60/6 = 10 A. As a check, (140-60)/20 = 4 A flows in from the left and (90-60)/5 = 6 A flows in from the right, and 4 + 6 = 10. The answer is (c).

The idea to keep. When two batteries feed a shared middle resistor, ground one wire and write one junction equation. Solving for one node voltage is quicker than solving two loop equations.

Answer from the NTA final answer key. Solution by PrepWiser.