JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6Ω. Calculate the power dissipated in the whole circuit :
- (a) 1.32 W
- (b) 0.65 W
- (c) 2.2 W
- (d) 4.4 W
Answer and solution
Answer (c) 2.2 W
Solution
Physics sir · 4 steps
Follow each wire: all four branches run straight from A to B, so they are just four resistors in parallel.
- Find the nodes. The whole left side of the rectangle is one wire, so both left corners are point A. The whole right side is one wire, so both right corners are point B. The little hop in the middle means the two diagonals cross without touching.
- Four branches. Top: 4 Ω. Bottom: 8 Ω. The diagonal from top-left to bottom-right is 4 + 8 = 12 Ω. The diagonal from bottom-left to top-right is 2 + 4 = 6 Ω. Each one joins A to B.
1/R = 1/4 + 1/8 + 1/12 + 1/6 = (6 + 3 + 2 + 4)/24 = 15/24 ⇒ R = 1.6 Ω
- Current from the cell. The total resistance is 1.6 + 0.6 = 2.2 Ω, so I = 2.2/2.2 = 1 A.
- So. Power dissipated in the whole circuit, internal resistance included, is P = EI = 2.2 × 1 = 2.2 W. That is option (c).
The idea to keep. Before you simplify any network, mark which points are the same node. A wire hop means no junction. For the whole circuit use P = EI, which already counts the cell's internal resistance.
Answer from the NTA final answer key. Solution by PrepWiser.