Page 23 - 12-phy-13 Current Electricity

13. Current Electricity eLearn.Punjab

The convention regarding the sign of the current is that if we are applying the Kirchhoff’s second
rule in the first loop, then the current of this loop i.e. I will be assigned a positive sign and all
1
currents, flowing opposite to I have a negative sign. Similarly, while applying Kirchhoff’s second
1
rule in the second loop, the current I will be considered as positive and I as negative. Using this
1
2
convention the current flowing through R is (I - I ) and the voltage change across is - (I - I ) R . The
2
2
1
2
2
1
voltage change across the battery E is E . Thus the Kirchhoff’s second rule as applied to the loop
2
2
abcda gives
) + =0
-IR
− E 1 11 − ( − I 1 I R E 2
2
2
Substituting the values, we have
( − I
40V − I 1 x10W− I 1 2 )x30W+ 60V = 0
−
20V 10 x  − W 1  3( − I 1 2 ) + I  = I 0
I
-1
or 4 − I 2VW= 2A .....(13.21)
3 =
1 2
Similarly applying Kirchhoff's second rule to the loop ebcfe, we get
− 2 − E ( − I 2 I 1 ) -R IR + =0E 3
2
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Substituting the values, we have
60V (− − I 2 − I 1 )x30W− I 2 x15W+ 50V = 0

10V 15 x− − W 2  + I  2( 2 1 ) − I  = I 0

-1
or 6 −I 9 =I 2VW= 2A .....(13.22)
1 2
I
Solving Eq. 13.21 and Eq.13.22 for and ,we get
I
1
2
2 2
I
I
= 3 A and = 9 A
1
2
I
Knowing the values of loop currents and I the actual current flowing through
1 2
each resistance of the circuit can be determined. Fig. 13.22 shows that and I I
1 2
are the actual currents through the resistances and . The actual current R 1 R 3
I
through R is the difference of and and its direction is along the larger current.
I
2 1 2
Thus
2
The current through R = I = A = 0.66A flowing in the direction of i.e. from a to d.
I
1 1 3 1
2 2
The current through R 2 = −I 1 I = 3 A- A = 0.44A flowing in the direction of i.e. fromI 1
2
9
c to b.

2
The current through R = I = A = 0.22A flowing in the direction of i.e. from f to e.I
3 2 9 1

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