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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 answers. 4. Do you expect the current in R2 to remain unchanged if Ea and Eb were transposed? Check your answer by experiment. 3