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Transcript
```Memorial University of Newfoundland
Department of Physics and Physical Oceanography
Physics 2055 Laboratory
Kirchhoff’s Circuit Laws
Introduction
Kirchhoff’s laws (named after Gustav Robert Kirchhoff, 1824–1887) are extensions of Ohm’s law
and enable the current in any part of a network to be calculated. The two laws may be summarized
as follows:• The algebraic sum of currents at a junction is zero, i.e. the sum of the currents leaving a
junction is equal to the sum of the currents arriving at that junction.
• For any closed loop, the algebraic sum of the voltage drops is equal to the algebraic sum of
the emfs. This is a generalization of Ohm’s law and may be written as
∑ IR = ∑ E .
Applying the first law to node B in Fig (1), we get
I1 = I2 + I3 ,
and applying the second law to the loop ABEF
Ea = I1 R1 + I2 R2 .
1. Construct the circuit as shown using three different resistors (each <
∼ 1000 Ω) for R1 , R2 and
R3 . Ea and Eb should be approximately 5 volts and 12 volts, respectively. Measure the
resistances and voltages as accurately as you can. How large are the uncertainties which are
associated with each of these measurements?
2. Measure the current in each resistor. What are the uncertainties associated with each of the
measured currents? Copy the circuit diagram into your book, indicating the correct directions
of current.
1
A
R1
R3
B
I1
C
I3
εa
R2
εb
I2
F
E
D
Figure 1: A network to study Kirchhoff’s laws
3. Use the second law on loops BCDE and ACDF to obtain two more equations for Ea and Eb
in terms of the various currents and resistances, and solve these equations to calculate I1 , I2
and I3 . How well do these currents agree with your measured values?
4. We have ignored the internal resistance of the power supply. How might you determine its
internal resistance? Do you think it should be included in your calculations? If not, why
not?
Principle of Superposition
The principle of superposition states that “the total current in any part of a linear network is equal
to the algebraic sum of the currents which would exist in that part if each of the emfs in turn were
acting alone, the other sources being replaced by their internal resistance.”
For the purpose of this experiment we shall assume the internal resistance of each source to be
negligible.
1. Measure the voltage drop across each resistor, V1 , V2 and V3 .
2. Remove the 5 volt supply, connect node A to ground and measure the voltage drops V10 , V20
and V30 across R1 , R2 and R3 , respectively.
3. Reconnect the source, and remove the 12 volt supply, connecting node C to ground. Similarly, measure V100 , V200 and V300 and show by experiment that
V1 = V10 +V100
2
V2 = V20 +V200
V3 = V30 +V300
You will need to unambiguously identify the signs of each of the voltages to obtain correct