JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

Two resistors of 100 $Ω$ each are connected in series with a 9V battery. A voltmeter of 400 $Ω$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is____V.

A
$3$
B
$4.5$
C
$4$
D
$2$

Solution

The voltmeter (400 $Ω$ ) sits in parallel with one of the 100 $Ω$ resistors. Their parallel combination is

$R_{p} = \frac{100 \times 400}{100 + 400} = 80 Ω$ .

Total circuit resistance: $R_{e q} = 100 + 80 = 180 Ω$ .

Current from the battery: $I = \frac{9}{180} = \frac{1}{20}$ A.

Voltmeter reads the voltage across the parallel combination:

$V = I \times R_{p} = \frac{1}{20} \times 80 = 4$ V.

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