Question 24.14: In Fig. 24.19, let R1 = 2 Ω, R2 = 4 Ω, E1 = 6 V, E2 = 3 V, a......
In Fig. 24.19, let R_{1} = 2 Ω, R_{2} = 4 Ω, \mathcal{E}_{1} = 6 V, \mathcal{E}_{2} = 3 V, and C = 2 µF. Find the steady currents I_{1}, I_{2}, and I_{3} and the charge Q.
Applying Kirchhoff’s junction rule at point b, we get:
(1) Junction b: I_{1}=I_{2}+I_{3}
The application of the loop rule to the loops abgha and bcfgb gives:
Loop abgha: -\,I_{1}\,R_{1}-{\mathcal{E}}_{2}-I_{2}\,R_{2}-I_{1}\,R_{1}+{\mathcal{E}}_{1}=0
(2) -\,4\,I_{1}-4\,I_{2}+3=0
Loop bcfgb: -\,I_{3}\,R_{1}-\mathcal{E}_{1}-I_{3}\,R_{1}+I_{2}\,R_{2}+\mathcal{E}_{2}=0
-2\,{ I}_{3}-6-2\,{ I}_{3}+4\,{I}_{2}+3=0(3) -4I_{3}+4I_{2}-3=0
Substituting Eqs. (1) into (3) gives:
(4) -4I_{1}+8I_{2}-3=0
Subtracting Eqs. (4) from (2) gives:
(-4\,I_{1}-4\,I_{2}+3)-(-4\,I_{1}+8\,I_{2}-3)=0\quad\Rightarrow\quad I_{2}=0.5\,\mathrm{A}Using this value of I_{2} in Eq. (4) gives I_{1} a value of:
-4\ I_{1}+8(0.5)-3=0\quad\Rightarrow\quad I_{1}=0.25\;\mathrm{A}Finally, from Eq. (1) we have:
I_{3}=I_{1}-I_{2}=0.5-0.25\ \mathrm{A}=0.25\ \mathrm{A}Applying Kirchhoff’s loop rule to the loop cdefc gives:
Loop cdefc: Q/C-{\mathcal{E}}_{2}+{\mathcal{E}}_{1}=0\quad\Rightarrow\quad Q=({\mathcal{E}}_{2}-{\mathcal{E}}_{1})C=-6\,\mu C