Page 159 - Physics Coursebook 2015 (A level)

Chapter 10: Kirchhoff’s laws
WORKED EXAMPLE
2 Calculate the current in each of the resistors in the circuit shown in Figure 10.11.
Step1 Markthecurrentsflowing.Thediagram shows I1, I2 and I3.
Hint: It does not matter if we mark these flowing in the wrong directions, as they will simply appear as negative quantities in the solutions.
Step2 ApplyKirchhoff’sfirstlaw.AtpointP,thisgives: I1 + I2 = I3 (1)
Step3 ChoosealoopandapplyKirchhoff’ssecondlaw. Around the upper loop, this gives:
6.0 = (I3 × 30)+(I1 × 10) (2)
Step4 Repeatstep3aroundotherloopsuntilthereare the same number of equations as unknown currents. Around the lower loop, this gives:
We now have three equations with three unknowns (the three currents).
Step5 Solvetheseequationsassimultaneous equations. In this case, the situation has been chosen to give simple solutions. Equation 3 gives I3 = 0.067 A, and substituting this value in equation 2 gives I1 = 0.400 A. We can now find I2 by substituting in equation 1:
I2 = I3 − I1 = 0.067−0.400 = −0.333A ≈−0.33A
Thus I2 is negative – it is in the opposite direction to the arrow shown in Figure 11.11.
Note that there is a third ‘loop’ in this circuit; we could have applied Kirchhoff’s second law to the outermost loopofthecircuit.Thiswouldgiveafourthequation:
6−2=I1 ×10
However, this is not an independent equation; we
could have arrived at it by subtracting equation 3 from equation 2.
QUESTION
2.0= I3 ×30
E1 E2
I1 R1
I2 R2
(3)
I1 R4
I2 R3
6
You can use Kirchhoff’s second law to find the current I in the circuit shown in Figure 10.13. Choosing the best loop can simplify the problem.
a Which loop in the circuit should you choose?
b Calculate the current I.
5.0 V
20Ω
5.0 V
2.0 V
10Ω
E3
Figure 10.12 A loop extracted from a complicated circuit.
e.m.f.s
Starting with the cell of e.m.f. E1 and working anticlockwise around the loop (because E1 is ‘pushing current’ anticlockwise):
sum of e.m.f.s = E1 +E2 −E3
Note that E3 is opposing the other two e.m.f.s.
p.d.s
I
10Ω
Starting from the same point, and working anticlockwise again:
sum of p.d.s = I1R1 − I2R2 − I2R3 + I1R4
Note that the direction of current I2 is clockwise, so the p.d.s that involve I2 are negative.
Figure10.13 Carefulchoiceofasuitableloop can make it easier to solve problems like this. For Question 6.
5.0 V
147