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In the given circuit the internal resistance of the 18V cell is negligible. If R1 = 400\Omega, R3 = 100\Omega and R4 = 500\Omega and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be:

  • Option 1)

     450\Omega

  • Option 2)

    300 \Omega

  • Option 3)

    550\Omega

  • Option 4)

    230\Omega

Answers (1)

best_answer

 

Potentiometer -

It is a device used to measure e.m.f of a given cell and to compare e.m.f's of cells

- wherein

It is also used to measure internal resistance of given cell

 

Current through 500\Omega , I_{2}=\frac{5}{500}=\frac{1}{100}A

\therefore  Voltage across 100\Omega = I_{2}R

                                         = \frac{1}{100}\times 100

                                         = 1 V

So Voltage across R_{2}=5+1=6V

Voltage across R_{1}=18-6=12V

Current through R_{1}

I=\frac{12}{400}=\frac{3}{100}

Current through R_{2}

I_{1}=I-I_{2}

     =\frac{3}{100}-\frac{1}{100}

    =\frac{2}{100}

R_{2}=\frac{6}{\frac{2}{100}}=300\Omega


Option 1)

 450\Omega

Option 2)

300 \Omega

Option 3)

550\Omega

Option 4)

230\Omega

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