Current Electricity

JEE Main level, single correct, medium

69 of 380

Question 69 of 380

Two resistors R1=(4±0.8)Ω{R}_{1}=(4\pm 0.8)\Omega and R2=(4±0.4)Ω{R}_{2}=(4\pm 0.4)\Omega are connected in parallel. The equivalent resistance of their parallel combination will be :

Show answer and solution

Answer: (D) (2±0.3)Ω(2\pm 0.3)\Omega

ΔRR2=ΔR1R12+ΔR2R22\dfrac{\Delta R}{{R}^{2}}=\dfrac{\Delta {R}_{1}}{{R}_{1}^{2}}+\dfrac{\Delta {R}_{2}}{{R}_{2}^{2}}
ΔR=(R)2[ΔR1R12+ΔR2R22]\Delta R={\left(R\right)}^{2}\left[\dfrac{\Delta {R}_{1}}{{R}_{1}^{2}}+\dfrac{\Delta {R}_{2}}{{R}_{2}^{2}}\right]
=(2)2[0.842+0.442]={\left(2\right)}^{2}\left[\dfrac{0.8}{{4}^{2}}+\dfrac{0.4}{{4}^{2}}\right]
=[0.2+0.1]=0.3=[0.2+0.1]=0.3
Rnet=R+ΔR=2±0.3{R}_{\text{net}}=R+\Delta R=2\pm 0.3

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Circuit builder simulationDrag the sliders and watch the numbers in this question change.Revise Current Electricity firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Current Electricity (Resistance and Resistivity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.