Electrical circuits - Kirchhoff's laws etc.

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Jmun
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#1
Report Thread starter 5 years ago
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Attached is the circuit diagram.

a) Use Kirchhoff's second law for loop DEHG to show that I2 = 4 I3

This was easy;
e.m.f.'s = p.d.'s
V = IR
V2 = I2 x 2 = 2 I2
V3 = I3 x 8 = 8 I3
2 I2 = 8 I3
I2 = 4 I3

b) Hence calculate the values I2 and I3

2.0 A = I2 + I3 = 4 I3 + I3 = 5 I3
I3 = 2/5 = 0.4 A
I2 = 4 I3 = 4 x 0.4 = 1.6 A

c) Use Kirchhoff's first law to find I1

Kirchhoff's first law is " The sum of the currents entering a junction in a circuit is equal to the currents leaving it." I'm stick here.

d) To find the e.m.f. of the battery you need to find a suitable loop around which to apply Kirchhoff's second law.

(i) State which loop you have chosen

Here, I don't know which loop to choose...

(ii) Calculate the e.m.f

Thanks
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Joinedup
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c) you have a choice of looking at either D or G

looking at D
1=1.6+i1
i1=1-1.6
i1=-0.6

looking at G
1+i1=0.4
i1=0.4-1
i1=-0.6

NB the current doesn't have to be flowing in the same direction as the arrow is pointing, so it is allowed to be +ve or -ve

---

any loop including the battery will do... so you might as well choose one with fewest resistors.
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Jmun
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Thanks! Great.

For the loop, would BACDE be suitable?
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Jmun
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So, if i choose BACDE then use 2nd law, e.m.f.'s = p.d.'s
V = IR so,
e.m.f = (2 X 3) + (1 X 4) + (1.6 X 2) = 13.2 V
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