Question

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?

Answer 1

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

$\frac{V'' - V'}{V'' }\times 100 = \left ( \frac{22.73-26.8}{22.73} \right )\times 100$ = 18 %