Current Electricity

JEE Main level, numerical answer, medium

68 of 380

Question 68 of 380

A 220V,50Hz220V,50Hz AC source is connected to a 25V,5W25V,5W lamp and an additional resistance RR in series (as shown in figure) to run the lamp at its peak brightness, then the value of RR (in ohm) will be

Show answer and solution

Answer: 975

Resistance of the bulb can be calculated as,
P=V2RB⇒RB=V2P=2525=125ΩP=\dfrac{{V}^{2}}{{R}_{B}}\Rightarrow {R}_{B}=\dfrac{{V}^{2}}{P}=\dfrac{{25}^{2}}{5}=125\Omega
The current through the bulb for peak brightness should be,
i=25125=15Ai=\dfrac{25}{125}=\dfrac{1}{5}A
Now, irms=15=220RB+R{i}_{rms}=\dfrac{1}{5}=\dfrac{220}{{R}_{B}+R}
RB+R=1100RB+R=1100
R=1100−125=975R=1100-125=975

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.