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

Given: Two resistors of 100 each are in series with a 9 V battery. A voltmeter of resistance 400 is connected across one 100 resistor. Find: The voltmeter reading. The voltmeter is connected in parallel with one 100 resistor. R parallel = 100 400100 + 400 = 80 The total circuit resistance becomes R total = 100 + 80 = 180 Current drawn from the battery is I = 9180 = 0.05 A The voltage across the parallel branch, which is the voltmeter reading, is V = I 80 = 0.05 80 = 4 V Therefore, the voltmeter reads 4 V. The correct option is C.

Two resistors of 100 each are connected in series with a 9 V batt… — Solution | JEE Physics

MCQEasyJEE 2026Kirchhoff's Laws & Circuits

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

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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two resistors of $100\,\Omega$ each are in series with a $9\,V$ battery. A voltmeter of resistance $400\,\Omega$ is connected across one $100\,\Omega$ resistor.

Find: The voltmeter reading.

The voltmeter is connected in parallel with one $100\,\Omega$ resistor.

R_{\text{parallel}} = \dfrac{100 \times 400}{100 + 400} = 80\,\Omega

The total circuit resistance becomes

R_{\text{total}} = 100 + 80 = 180\,\Omega

Current drawn from the battery is

I = \dfrac{9}{180} = 0.05\,\text{A}

The voltage across the parallel branch, which is the voltmeter reading, is

V = I \times 80 = 0.05 \times 80 = 4\,\text{V}

Therefore, the voltmeter reads $4\,\text{V}$ . The correct option is C.

Common mistakes

Treating the voltmeter as ideal and ignoring its finite resistance is incorrect because the $400\,\Omega$ voltmeter changes the branch resistance. Instead, first take the parallel combination of $100\,\Omega$ and $400\,\Omega$ .
Adding $100\,\Omega$ and $400\,\Omega$ directly for the measured branch is wrong because they are connected in parallel, not in series. Use the parallel resistance formula to get $80\,\Omega$ .
Using the total current and multiplying it by $100\,\Omega$ for the reading is incorrect because the measured branch no longer has resistance $100\,\Omega$ alone. The branch voltage must be found across the equivalent parallel resistance $80\,\Omega$ .

Practice more Kirchhoff's Laws & Circuits questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Kirchhoff's Laws & Circuits

Answer

Step-by-step solution

Common mistakes

Practice more Kirchhoff's Laws & Circuits questions

Related questions