Question 121 of 380

A cell of emf $90V$ 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:

A$40V$B$45V$C$80V$D$90V$

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Answer: (A) $40V$

As we know, voltmeter is connected in parallel to the element being measured.
The equivalent resistance of the above circuit is ${R}_{eq}=\left(\dfrac{400\times 100}{400+100}\right)+100=80+100=180\Omega$
Current through the circuit is $i=\dfrac{90}{180}=\dfrac{1}{2}A$
Reading of the voltmeter is ${V}_{R}=i\left(\dfrac{400\times 100}{400+100}\right)=\dfrac{1}{2}\times 80=40V$