Voltmeter range

We have a voltmeter that, in its original setting, measures up to 10 V. Calculate the resistance of the series resistor (multiplier) needed to extend the range to 50 V. The voltmeter's internal resistance is 2 kΩ/V.

Final Answer:

R₂ =  80000 Ω

Step-by-step explanation:

$$\begin{gathered} U_1=10\text{V} \\ {U}_2=50\text{V} \\ {R}_1=20\cdot 1000=20000\Omega \\ I=\frac{U_1}{R_1}=\frac{10}{20000}=\frac{1}{2000}=0.0005\text{A} \\ {U}_3=U_2-U_1=50-10=40\text{V} \\ {U}_3=I\cdot R_2 \\ {R}_2=U_3/I=40/\frac{10}{20000}=40:\frac{10}{20000}=40\cdot \frac{20000}{10}=\frac{40\cdot 20000}{10}=\frac{800000}{10}=80000\Omega \end{gathered}$$

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