Current Electricity

JEE Main level, single correct, medium

158 of 380

Question 158 of 380


A
9V9V battery with an internal resistance of 0.5Ω0.5\Omega is connected across an infinite network, as shown in the figure. All ammeters A1,A2,A3{A}_{1},{A}_{2},{A}_{3} and voltmeter VV are ideal.
Choose the correct statement.

Show answer and solution

Answer: (D) Reading of A1{A}_{1} is 2A2A.

In the given infinite network problems, let, XX be an equivalent resistance. If we connect one section of an infinite network with the replacing network using an equivalent resistance, then in this new equivalent circuit the equivalent resistance will still remain the same.
Let,
Req=x{R}_{eq}=x
x=4x4+x+2x=\dfrac{4x}{4+x}+2
x=8+6x4+xx=\dfrac{8+6x}{4+x}
4x+x2=8+6x4x+{x}^{2}=8+6x
x2−2x−8=0{x}^{2}-2x-8=0.
On solving,

x=4,x=−2x=4,x=-2
Here, the negative value of resistance is not possible.

So, resistance is
x=4Ωx=4\Omega
Reading of
A1{A}_{1} is I=VX+r...(1)I=\dfrac{V}{X+r}...(1),
where,
V=9Volt,r=0⋅5Ω,R=4ΩV=9Volt,r=0\cdot 5\Omega ,R=4\Omega.
Then, from the above equation,

I=94+0.5=2AI=\dfrac{9}{4+0.5}=2A.

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 (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.