A voltage divider consisting of two 4.7-k Ohm resistors is connected in series t

Question

A voltage divider consisting of two 4.7-k Ohm resistors is connected in series to a 60-V battery with a negligible internal resistance. You want to measure the potential difference across one of the resistors using a voltmeter having its own resistance of 10 k Ohm. What is the magnitude of the error introduced by this voltmeter (the difference between the reading of the voltmeter and the real voltage across the 4.7 k Ohm resistor before the voltmeter was attached)? 5.71 V 4.06 V 1.17 V 2.58 V 3.15 V

Explanation / Answer

The real voltage on the R0 = 4.7k resistor before the voltmeter was attached is,

V0 = 60 V/(R1 + R2) = [(60 V)/(4.7k + 4.7k)]4.7k

or, V0 = 30 V

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When the voltmeter with resistance R = 10 k is attached to resistor R0, the equivalent resistance of the circuit becomes,

Req = R ll R0 + R0 = (4.7kX10k)/14.7k + 4.7k

or, Req = 7.897 k

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So current given by the battery is,

I = 60V/ 7.897 k

or, I = 7.597A

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This current is divided among R0 and R. So the current through the voltmeter is

I' = [4.7k/(4.7k + 10 k)]7.597A

or, I' = 2.429 A

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So reading on the voltmeter will be,

Vm = I'R = (2.429 A)(10 k)

or, Vm = 24.29 V

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So difference is reading is,

V = 30 V - 24.29 V

or, V = 5.71 V

So, a) is the correct option

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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