The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

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2 years ago

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