JEE Main 2021 · 18 March · morning · Numerical answer · +4 / −0
The voltage across the 10Ω resistor in the given circuit is x volt.
The value of 'x' to the nearest integer is _________.
Answer and solution
Answer 70
Solution
Physics sir · 3 steps
The 50 and 20 ohm resistors share both ends, so combine them first, then one loop gives the current.
- Spot the parallel pair. The 50 Ω and 20 Ω resistors start at the same junction and end on the same right-hand wire, so they are in parallel. Their combination is (50 × 20)/(50 + 20) = 100/7 Ω.
- Apply the loop law. This combination is in series with the 10 Ω resistor and the battery. Going once around the loop, the drops must add up to the 170 V supplied.
170 = I(10 + 100/7) = I × 170/7 ⇒ I = 7 A
- So. The full 7 A flows through the 10 Ω resistor, so the voltage across it is 7 × 10 = 70 V, giving x = 70.
The common slip: 100. Reporting the voltage across the parallel pair, 7 × 100/7, gives 100, not 70. The question asks for the drop across the 10 Ω resistor.
The idea to keep. Before writing Kirchhoff equations, look for resistors that share both end points. Reducing them first often turns the circuit into one loop with one current.
Answer from the NTA final answer key. Solution by PrepWiser.