Two cells of emf 13 V and 12 V and negligible internal resistance are connected in the circuit…
Two cells of emf $13\text{ V}$ and $12\text{ V}$ and negligible internal resistance are connected in the circuit shown in Figure 8 below.
Using Kirchhoff's laws, calculate the current flowing through $5\ \Omega$ resistor.
Let $I_1$ flow in the 13 V branch and $I_2$ in the 12 V branch; both reach D and pass through the $5\ \Omega$ resistor, so the current in it is $I_1 + I_2$.
Loop BADCB (Kirchhoff's loop rule): $13 - 3I_1 - 5(I_1 + I_2) = 0$, i.e. $8I_1 + 5I_2 = 13$.
Loop BDCB: $12 - 2I_2 - 5(I_1 + I_2) = 0$, i.e. $5I_1 + 7I_2 = 12$.
Solving: $I_1 = 1\text{ A}$, $I_2 = 1\text{ A}$, so the current through $5\ \Omega$ is $2\text{ A}$ (from D to C).
