Question 144 of 380

Two resistors $400\Omega$ and $800\Omega$ are connected in series across a $6∨$ battery. The potential difference measured by a voltmeter of $10k\Omega$ across $400\Omega$ resistor is close to:

A$2V$B$1.8V$C$2.05V$D$1.95V$

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Answer: (D) $1.95V$

Let voltmeter reading is $v$
$\dfrac{v}{100}\times 400+\left(\dfrac{v}{10000}+\dfrac{v}{400}\right)800=6$
$\Rightarrow v+\dfrac{8v}{100}+2v=6;\dfrac{77v}{25}=6;v=\dfrac{150}{77}=1.95v$