Practice portal › Current Electricity › Kirchhoff's Laws and Circuit Analysis

The value of current in the 6 Ω resistance is [Figure: a 140 V battery on the left and a 90 V battery on the right, both with positive terminal up and negative terminal on the common bottom wire. From the top of the 140 V battery a 20 Ω resistor leads to a top junction X; from X a 5 Ω resistor leads to the top of the 90 V battery; a 6 Ω resistor joins X to the bottom wire.]

Asked in JEE Main 20th July 1st Shift 2021 · Kirchhoff's laws in networks

• (1) 8 A
• (2) 6 A
• (3) 4 A
• (4) 10 A

Step-by-step solution

Why the other options are wrong

• (1) Keeps only the 90 V source with 5 Ω and 6 Ω, dropping the 140 V branch: (90)/(11)≈8 A.
• (2) No standard slip gives this; it would need X at 36 V.
• (3) This is the current in the 20 Ω resistor, not in the 6 Ω.

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.