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Physics20183 marksNumericalCurrent ElectricityTool: Circuit solver (Kirchhoff's laws)

Solve the following.

ε₁ and ε₂ are two batteries having emf of 34V and 10V respectively and internal resistance of 1 Ω…

$\varepsilon_1$ and $\varepsilon_2$ are two batteries having emf of $34\text{V}$ and $10\text{V}$ respectively and internal resistance of $1\,\Omega$ and $2\,\Omega$ respectively. They are connected as shown in Figure 3 below. Using Kirchhoff’s Laws of electrical networks, calculate the currents $I_1$ and $I_2$.
Figure for this question

Answer

Answer

Applying Kirchhoff's laws to the network (current in the 5 $\Omega$ resistor is $I_1 + I_2$ by the junction rule): Loop FABEF (loop rule): $34 = I_1(1 + 4 + 7) + 5(I_1 + I_2)$, i.e. $17 I_1 + 5 I_2 = 34$ ... (1) Loop BCDEB (loop rule): $10 = I_2(2 + 4 + 7) + 5(I_1 + I_2)$, i.e. $5 I_1 + 18 I_2 = 10$ ... (2) Solving (1) and (2): $I_1 = 2\text{ A}$ and $I_2 = 0$.

Final answer: $I_1 = 2\text{ A},\ I_2 = 0$ A