What must be the emf E in the circuit so that the current flowing through the 7 Ω resistor is 1.80 A? Each emf source has negligible internal resistance.

What must be the emf E in the circuit so that the current flowing through the 7 Ω resistor is 1.80 A? Each emf source has negligible internal resistance.

Solution:
Given,
Current through 7 Ω resistor, $I_3$ = 1.80 A

Using Kirchhoff's junction rule,
At junction a,
$$\begin{array}{rl}{I}_1+I_2& =I_3\\ I_1+I_2& =1.80\text{... (i)}\end{array}$$ Using Kirchhoff's loop rule,
In loop ecdfe,
$$\begin{array}{rl}-I_3\times R_3-I_1\times R_1+E_1& =0\\ -1.80\times 7-I_1\times 3+24& =0\\ -12.6-3I_1+24& =0\\ 3I_1& =11.4\\ \therefore I_1& =3.8A\end{array}$$ From equation (i), $$I_2=1.80-3.8=-2A$$ In loop eabfe,
$$\begin{array}{rl}-E+I_2\times R_2-I_1\times R_1+E_1& =0\\ -E+(-2)\times 2-3.8\times 3+24& =0\\ -E-4-11.4+24& =0\\ \therefore E& =8.6V\end{array}$$ Hence, the emf E is 8.6 V .