Download Chapter 25 Problem 51 † Solution a) Find the voltage across the 30

Chapter 25
Problem 51
†
Rm
30 k W
100 V
40 k W
40 kW
Solution
a) Find the voltage across the 30 kΩ resistor using a 50 kΩ voltmeter.
Since the voltmeter is in parallel with the 30 kΩ resistor, the combination has a resistance of
1
1
1
=
+
R1
30 kΩ 50 kΩ
1
5
3
8
=
+
=
R1
150 kΩ 150 kΩ
150 kΩ
150 kΩ
R1 =
= 18.75 kΩ
8
The other two resistors are in parallel and have a combined resistance of
1
2
1
1
1
+
=
=
=
R2
40 kΩ 40 kΩ
40 kΩ
20 kΩ
R2 = 20 kΩ
The first combined resistance is in series with the second combined resistance as illustrated below
R1
100 V
R2
The total resistance for the circuit is then
RT = R1 + R2 = 18.75 kΩ + 20 kΩ = 38.75 kΩ
Using Ohm’s law, the current is
I=
V
100 V
=
= 2.5806 × 10−3 A = 2.5806 mA
3
RT
38.75 × 10 Ω
The voltage across the first combined resistance, which is also the voltage measured across the 30 kΩ
resistor is
V30kΩ = I · R1 = (2.5806 × 10−3 A)(18.75 × 103 Ω) = 48.39 V
b) Find the voltage across the 30 kΩ resistor using a 250 kΩ voltmeter.
Using the procedure for part a) the following results are obtained
1
1
1
28
=
+
=
R1
30 kΩ 250 kΩ
750 kΩ
†
Problem from Essential University Physics, Wolfson
R1 = 26.79 kΩ
RT = 26.79 kΩ + 20 kΩ = 46.79 kΩ
100 V
= 2.1372 × 10−3 A = 2.1372 mA
I=
46.79 × 103 Ω
V30kΩ = I · R1 = (2.1372 × 10−3 A)(26.79 × 103 Ω) = 57.26 V
c) Find the voltage across the 30 kΩ resistor using a 10 M Ω voltmeter.
Using the procedure for part a) the following results are obtained
1
1
1003
1
=
+
=
R1
30 kΩ 10000 kΩ
30000 kΩ
R1 = 29.91 kΩ
RT = 29.91 kΩ + 20 kΩ = 49.91 kΩ
100 V
= 2.004 × 10−3 A = 2.004 mA
I=
49.91 × 103 Ω
V30kΩ = I · R1 = (2.004 × 10−3 A)(29.91 × 103 Ω) = 59.94 V
If the meter were not in the circuit at all, the voltage would be 60 V . An ideal voltmeter should have an
infinite resistance so it has no effect on the circuit it is measuring.