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Electrical circuits - Kirchhoff's laws etc.
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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
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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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.
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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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
V = IR so,
e.m.f = (2 X 3) + (1 X 4) + (1.6 X 2) = 13.2 V
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