Physics Circle home Student Login

Home›Chapters›Current Electricity PYQs›Combination of Resistors›Question 50

JEE Main 2021 (17 Mar, Shift 1) · Combination of Resistors

Combination of Resistors question from JEE Main 2021 (17 Mar, Shift 1)

Q50JEE Main 2021 · 17 MAR · SHIFT 1Medium

Combination of Resistors

The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. n=sp=(R1+R2)2R1R2n = \dfrac{s}{p} = \dfrac{(R_1 + R_2)^2}{R_1 R_2}.
  2. Since (R1+R2)2≥4R1R2(R_1 + R_2)^2 \ge 4R_1R_2, n≥4n \ge 4, with equality for R1=R2R_1 = R_2.
  3. Minimum n=4n = 4
Report a mistake

More questions on Combination of Resistors