JEE Main 2021 · 16 March · morning · Numerical answer · +4 / −0
In the figure given, the electric current flowing through the 5 kΩ resistor is 'x' mA.
The value of x to the nearest integer is ____________.
Answer and solution
Answer 3
Solution
Physics sir · 3 steps
Collapse the three equal 3 kΩ resistors into one, then the whole circuit is a single loop.
- Parallel group. The three 3 kΩ resistors share the same two end nodes, so they are in parallel. Three equal resistors in parallel give 3/3 = 1 kΩ.
- Total resistance. There is now one loop: the 5 kΩ resistor, the 1 kΩ combination and the cell's internal resistance of 1 kΩ, all in series. That totals 5 + 1 + 1 = 7 kΩ.
I = E/Rₜₒₜₐₗ = (21 V)/(7 kΩ) = 3 mA
- So. The 5 kΩ resistor is in series with the cell, so the full loop current flows through it. That gives x = 3.
The common slip: 4. Leaving out the 1 kΩ internal resistance gives 21/6 = 3.5 mA, which rounds to 4 instead of 3.
The idea to keep. Reduce parallel groups first, then add everything in series, including the cell's internal resistance. A resistor in series with the cell carries the full loop current.
Answer from the NTA final answer key. Solution by PrepWiser.