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