Question 24.13: In Fig. 24.18, let R1 = 2 Ω, R2 = 6 Ω, R3 = 4 Ω, E1 = 10 V, ......
In Fig. 24.18, let R_{1} = 2 Ω, R_{2} = 6 Ω, R_{3} = 4 Ω, \mathcal{E}_{1} = 10 V, and \mathcal{E}_{2} = 14 V. Find the currents I_{1}, I_{2}, and I_{3} in the circuit.
We cannot simplify the circuit by the rule of adding resistances in series and in parallel. Thus, we must use Kirchhoff’s rules. By applying Kirchhoff’s junction rule to the junction f, we get:
(1) Junction f : I_{1}=I_{2}+I_{3}
We have three loops in this circuit, but we need only two loop equations to determine the three unknown currents. Applying Kirchhoff’s loop rule to the loops abcfa and fcdef and traversing these loops clockwise, we obtain the following equations (after temporarily omitting the units, since they are all consistent SI units):
(2) Loop abcfa: I_{1}R_{1}+I_{2}R_{2}-\mathcal{E}_{1}=0\quad\Rightarrow\quad2I_{1}+6I_{2}-10=0
(3) Loop fcdef : {\mathcal{E}}_{1}-I_{2}R_{2}+I_{3}R_{3}+{\mathcal{E}}_{2}=0\quad\Rightarrow\quad{24-6\,I_{2}+4\,I_{3}}=0
Substituting Eqs. (1) into (3) gives:
24 − 10 I_{2} + 4 I_{1} = 0
Dividing this equation by 2 gives:
(4) 12 − 5 I_{2} + 2 I_{1} = 0
Subtracting Eqs. (4) from (2) gives:
(2 I_{1} + 6 I_{2} − 10) − (12 − 5 I_{2} + 2 I_{1}) = 0 ⇒ I_{2} = 2 A
Using this value of I_{2} in Eq. (4) gives I_{1} a value of:
12 − 5 × 2 + 2 I_{1} = 0 ⇒ I_{1} = −1 A
Finally, from Eq. (1) we have:
I_{3} = I_{1} − I_{2} = −1 A − 2 A = −3 A
Thus ( I_{1} = −1A, I_{2} = 2A, I_{3} = −3A)
We notice that I_{1} and I_{2} are both negative. This means that the currents are opposite to the direction we chose. However, the numerical values are correct.