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,
\[\begin{align*} I_1+I_2&=I_3\\ I_1+I_2&=1.80 \hspace{0.1cm} \text {... (i)}\\ \end{align*}\] Using Kirchhoff's loop rule,
In loop ecdfe,
\[\begin{align*} -I_3 \times R_3 - I_1 \times R_1 + E_1 &=0\\ -1.80 \times 7 - I_1 \times 3 +24 &=0\\ -12.6 - 3I_1 +24&=0\\ 3I_1&=11.4\\ \therefore I_1&=3.8 \hspace{0.1cm} A \\ \end{align*}\] From equation (i), \[I_2=1.80-3.8=-2 \hspace{0.1cm} A\] In loop eabfe,
\[\begin{align*} -E+I_2\times R_2 - I_1 \times R_1 + E_1&=0\\ -E+(-2) \times 2 - 3.8 \times 3 +24 &=0\\ -E-4-11.4+24&=0\\ \therefore E&=8.6 \hspace{0.1cm} V\\ \end{align*}\] Hence, the emf E is 8.6 V .