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PHY 122 Circuit Solutions
Resistivity Problem An experiment is conducted to measure the electrical resistivity of nichrome in the
form of wires with different lengths and cross-sectional areas. For one set of measurements, a student uses
a 30-gauge wire, which has a cross-sectional area of 7.30 x 10-8 m2. The student measures the potential
difference across the wire and the current in the wire with a voltmenter and an ammeter, respectively.
•
•
For each of the measurements given
in the table taken on wires of three
different lengths, calculate the
resistance of the wires and the
corresponding values of the
resistivity.
L (m)
∆V (V)
I (A)
R (Ω
Ω)
ρ (Ω
Ωm))
0.540
5.22
0.500
10.44
1.411x10-6
1.028
5.82
0.276
21.09
1.498x10-6
1.543
5.94
0.187
31.76
1.503x10-6
What is the average value of the
resistivity, and how does this value compare with the value given for copper in the table in your text.
Avg ρnichrome = 1.47x10-6
ρcopper = 1.68x10-8 Much smaller (100 times) than nichrome!
R5
R2
V6
R6
V4
R3
R7
R1
R4
V1
V3
V2
R1=3 kΩ
R2=3 kΩ
R3=4 kΩ
R4=12 kΩ
R5=6 k
R6=6 k
R7=2 k
V= 24 V
+
Circuit Problem: Use the circuit shown answer the following questions. You will need to make a large copy of the
circuit to label all the values you find for parts (d) and (e) below.
a) Combine all parallel and series resistors to find an equivalent resistor.
b) Draw the equivalent circuit. Label the equivalent resistor with + and a - to indicate high and low potential.
c) Solve for the total current in the circuit.
d) Assume that the potential at the negative battery terminal is 0V and find the potential at all positions labeled V#.
Label these values on the circuit diagram.
e) Find the current that flows through each resistor. Label it with an arrow indicating the direction of current flow.
Req = 5.73 kΩ
Itotal = 4.19 mA
V1 = 24 V
V2 = 11.43 V
V3 = 4.57 V
V4 = 5.72 V
V6 = 0 V
I1 = 4.19 mA
I2 = 1.905 mA
I3 = 1.715 mA
I4 = 0.57 mA
I5 = 0.95 mA
I6 = 0.95 mA
I7 = 2.29 mA