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Two resistors $400\Omega$ and $800 \Omega$ are connected in series with a $\mathrm{6 V}$ battery. The potential difference measured by voltmeter of $\mathrm{10 k \Omega }$ across $\mathrm{400\Omega }$ resistor is:
Option: 1
$\mathrm{2\: V}$

Option: 2
$\mathrm{1.95\: V}$

Option: 3
$\mathrm{3.8\: V}$

Option: 4
$\mathrm{4\: V}$

Answers (1)

best_answer

Here, the resistances of $\mathrm{400 \Omega \: and \: 10000 \Omega}$ are in parallel, their effective resistance $\mathrm{R_P}$ will be

$\mathrm{R}_{\mathrm{P}}=\frac{400 \times 10000}{400+10000}=\frac{5000}{13} \Omega$

Total resistance of the circuit $=\frac{5000}{13}+800=\frac{15400}{13} \Omega$

Current in the circuit $\mathrm{I=\frac{6}{15400 / 13}=\frac{39}{7700} \mathrm{~A}}$

Potential difference across voltmeter $\mathrm{=\mathrm{IR}_{\mathrm{P}}}$

$\mathrm{ =\frac{39}{7700} \times \frac{5000}{13}=1.95 \mathrm{~V} }$

Hence Option 2 is correct.

Posted by

jitender.kumar

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