The voltmeter shown in figure 32-E25 reads 18 V across the 50 - resistor- Find the resistance of the voltmeter-

Let resistance of the voltameter be R Ω R_1 = 50R/50 + R R_2 = 24 Ω Both are in series 30 = V_1 + V_2 ⇒ 30 = iR_1 + iR_2 ⇒ 30 iR_2 = iR_1 ⇒ iR_1 = 30 = 30/R_1 ...

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Byju's Answer

Standard XII

Physics

Voltmeter

The voltmeter...

Question

The voltmeter shown in figure (32-E25) reads 18 V across the 50 $Ω$ resistor. Find the resistance of the voltmeter.

figure (32-E25)

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Solution

Let resistance of the voltameter be R $Ω$

$R_{1} = \frac{50 R}{50 + R}$

$R_{2} = 24 Ω$

Both are in series

$30 = V_{1} + V_{2}$

$\Rightarrow 30 = i R_{1} + i R_{2}$

$\Rightarrow 30 - i R_{2} = i R_{1}$

$\Rightarrow i R_{1} = 30 = \frac{30}{R_{1} + R_{2}} R_{2}$

$\Rightarrow V_{1} = 30 (1 - \frac{R_{2}}{R_{1} + R_{2}})$

$V_{1} = 30 (\frac{R_{1} + R_{2} + R_{2}}{R_{1} + R_{2}})$

$V_{1} = 30 (\frac{R_{1}}{R_{1} + R_{2}})$

$18 = 30 (\begin{matrix} \frac{50 R}{(50 + R) [\frac{50 R}{(50 + R)} + 24]} \end{matrix})$

$= 30 [\frac{50 R}{50 R + 24 (50 + R)}]$

$18 = \frac{30 \times 50 \times R}{74 R + 1200}$

$R = 130 Ω$

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The voltmeter shown in figure (32-E25) reads 18 V across the 50 Ω resistor. Find the resistance of the voltmeter. figure (32-E25)

Let resistance of the voltameter be R Ω R1=50R50+R R2=24Ω Both are in series 30=V1+V2 ⇒30=iR1+iR2 ⇒30−iR2=iR1 ⇒iR1=30=30R1+R2R2 ⇒V1=30(1−R2R1+R2) V1=30(R1+R2+R2R1+R2) V1=30(R1R1+R2) 18=30(50R(50+R)[50R(50+R)+24]) =30[50R50R+24(50+R)] 18=30×50×R74R+1200 R=130Ω

The voltmeter shown in figure (32-E25) reads 18 V across the 50 $Ω$ resistor. Find the resistance of the voltmeter.

figure (32-E25)