The GATE Grind

GATE 2014 EE (EE1) – Question 40

Electric Circuits · Network solution methods: KCL, KVL, Node and Mesh analysis · 2 marks · Numerical answer

For the circuit shown, R = 25 + I/2 Ω, with current I in amperes. Find I (A).

Diagram for GATE 2014 EE (EE1) question 40

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 10 to 10

Explanation

Ohm’s law gives300=I(25+I/2). Rearranging yields I²+50I−600=0=(I−10)(I+60). The passive positive-resistance solution is **I=10A**, with R=30Ω. The other algebraic root−60Awould make R=−5Ωand is excluded by the intended resistor model.