A circuit consists of a series combination of 6.00−kΩ and 5.00−kΩ resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00−kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

a. What potential difference does the voltmeter measure across the 5.00−kΩ resistor?

b. What is the true potential difference across this resistor when the meter is not present?

c. By what percentage is the voltmeter reading in error from the true potential difference?

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Vespertilio Vespertilio · Answer 1 · 2020-03-28T13:10:38+01:00

Answer:

Explanation:

Resistance, R1 = 6 kilo ohm

Resistance, R2 = 5 kilo ohm

Resistance of voltmeter, R = 10 kilo ohm

Voltage , V = 50 V

(a) When the voltmeter is applied

R and R2 in parallel. Rp = 10 x 5/ 15 = 3.33 kilo ohm

Now, Rp and r1 in series,

Req = 6 + 3.33 = 9.33 kilo ohm

Let i is the total current

i = V/ Req

i = 50 / (9.33 x 1000) = 5.36 mA

Let potential difference across the 5 kilo ohm is V'.

V' = i x R2 = 5.36 x 10^-3 x 5 x 1000 = 26.8 V

(b) When the meter is not applied

R1 and R2 in series

Req = R1 + R2

Req = 6 + 5 = 11 kilo ohm

i = V / Req = 50 / 11 = 4.55 mA

Let the potential difference across the 5 kilo ohm is V''.

V'' = i x R2 = 4.55 x 10^-3 x 5 x 1000 = 22.73 V

(c) Percentage error

[tex]\frac{V'' - V'}{V'' }\times 100 = \left ( \frac{22.73-26.8}{22.73} \right )\times 100[/tex] = 18 %