A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

Question

A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be : (A) 45 V (B) 40 V (C) 80 V (D) 90 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/78be8ef164be120a583b1c71cda11da3-.png" /> R text eq =(400 × 100/500)+100 =180 Ω i =(90/180)=(1/2) A Reading =(1/2) × (400/500) × 100 =40 text volt

Q. A cell of emf $90 V$ is connected across series combination of two resistors each of $100 Ω$ resistance. A voltmeter of resistance $400 Ω$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

$45 V$

$40 V$

$80 V$

$90 V$

$R_{eq} = \frac{400 \times 100}{500} + 100$
$= 180Ω$
$i = \frac{90}{180} = \frac{1}{2} A$
Reading $= \frac{1}{2} \times \frac{400}{500} \times 100$
$= 40 volt$

Q. A cell of emf 90V is connected across series combination of two resistors each of 100Ω resistance. A voltmeter of resistance 400Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

45V

40V

80V

90V