Question

A 36 V battery of negligible internal resistance is connected to a 47 kΩ and a 21 kΩ resistor in series.

a).

What reading will a voltmeter, of internal resistance 95 kΩ , give when used to measure the voltage across the first resistor?

Express your answer using two significant figures.

V47 kΩ=

b).

What is the percent inaccuracy due to meter resistance for each case?

Express your answer using two significant figures.

V not ideal−V ideal/V ideal=

c).

What reading will a voltmeter, of internal resistance 95 kΩ , give when used to measure the voltage across the second resistor?

Express your answer using two significant figures.

V21 kΩ=

d).

What is the percent inaccuracy due to meter resistance for each case?

Express your answer using two significant figures.

V not ideal−V ideal/V ideal=

Answer 1

a)

with ideal conditions

equivalent resistance, R = 47 + 21 = 68 k

so,

I = V / R = 36 / 68e3

I = 5.294e-4 A

Now, find the ideal voltage across each resistor,

V (47) = 5.294e-4 * 47e3 = 24.8823 V

V(21) = 5.294e-4 * 21e3 = 11.1174 V

now, put the 47 k resistor in parallel with 95 k

we have

1/R = 1/47 + 1/95

R = 31.4436 k

so

R (total) = 31.4436 + 21 = 52.4436 k

so,

I = 36 / 52.4436e3

I = 6.8645e-4 A

so,

V (47) = 6.8645e-4 * 31.4436e3

V (47) = 22 Volts

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(b)

% error = (21.58 - 24.8823 / 24.8823) * 100

% error = - 13 %

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(c)

now, put the voltmeter in parallel with 21 k

1/R = 1/95 + 1/21

R = 17.198 k

R (total) = 17.198 + 47 = 64.198 k

I = 36 / 64.198e3

I = 5.607e-4 A

so,

V (21) = 5.607e-4 * 17.198e3

V (21) = 9.6 V

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(c)

% error = (9.6439 - 11.1174 / 11.1174 ) * 100

% error = -13 %