### Current Electricity Question 3

#### Question 3 - 24 January - Shift 2

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

(1) $40 V$

(2) $45 V$

(3) $80 V$

(4) $90 V$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Combination of Resistances

$R _{\text{eq }}=\frac{400 \times 100}{500}+100$

$=180 \Omega$

$i=\frac{90}{180}=\frac{1}{2} A$

Reading $=\frac{1}{2} \times \frac{400}{500} \times 100$