In the given circuit the internal resistance of the 18 V cell is negligible- If R1-400 - - R3-100 - and R4-500 - and the reading on an ideal voltmeter across R4 is 5 V- then the value of R2 will be

Reading of voltameter across R_4, V_4=5 V nbsp; So, current throgh R_4, i_1=V_4R_4=0.01 A Voltage across R_3, V_3=i_1R_3=0.01×100=1 V Total voltage V_3+V_4= ...

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Byju's Answer

Standard XII

Physics

Voltmeter

In the given ...

Question

In the given circuit the internal resistance of the $18 V$ cell is negligible. If $R_{1} = 400 Ω, R_{3} = 100 Ω$ and $R_{4} = 500 Ω$ and the reading on an ideal voltmeter across $R_{4}$ is $5 V$ , then the value of $R_{2}$ will be

230 Ω

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300 Ω

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450 Ω

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550 Ω

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Solution

The correct option is B $300 Ω$

Reading of voltameter across $R_{4}$ ,
$V_{4} = 5 V$
So, current throgh $R_{4}$ ,
$i_{1} = \frac{V_{4}}{R_{4}} = 0.01 A$
Voltage across $R_{3},$
$V_{3} = i_{1} R_{3} = 0.01 \times 100 = 1 V$
Total voltage
$V_{3} + V_{4} = 6 V = V_{2}$
From circuit, we can write-
$V_{1} + V_{3} + V_{4} = 18 V$
$\Rightarrow V_{1} = 12 V$
$i = \frac{V_{1}}{R_{1}} = 0.03 A$
Using KCL,

$\Rightarrow i = i_{1} + i_{2} \Rightarrow i_{2} = i - i_{1} = 0.03 - 0.01 A = 0.02 A$
$∴ R_{2} = \frac{V_{2}}{i_{2}} = \frac{6}{0.02} = 300 Ω$

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

In the given circuit the internal resistance of the $18 V$ cell is negligible. If $R_{1} = 400 Ω, R_{3} = 100 Ω$ and $R_{4} = 500 Ω$ and the reading on an ideal voltmeter across $R_{4}$ is $5 V$ , then the value of $R_{2}$ will be