Question

The figure shows a network of five resistance and two batteries. The current through the 15 V battery is :

• A. zero
• B. 1 A
• C. 3 A
• D. None of the above

Answer 1

The correct option is D None of the above

$\begin{array}{c}{V}_E+15-i-3i_{{}^{\prime }}-2i=VE\\ \Rightarrow 15-3i-3i_{{}^{\prime }}=0\\ \Rightarrow 15=3i+3i_{{}^{\prime }}\\ \Rightarrow 5=i+i_{{}^{\prime }}......\left(1\right)\\ V_E+15-i-4\left(i-i_{{}^{\prime }}\right)-30-2\left(i-i_{{}^{\prime }}\right)-2i=VE\\ \Rightarrow -15-i-4i+4i_{{}^{\prime }}-2i+2i_{{}^{\prime }}=0\\ \Rightarrow -15+6i_{{}^{\prime }}-7i=0\\ \Rightarrow 7i-6i_{{}^{\prime }}=\mathrm{15.........}\left(2\right)\\ Nowmultiplyequation\left(1\right)\times 6thenaddinequation\left(2\right)\\ \underset{–}{\begin{array}{c}6i+6i_{{}^{\prime }}=30\\ 7i-6i_{{}^{\prime }}=15\end{array}}\\ 13i=45\\ i=\frac{45}{13}\\ Hence,\\ optionDiscorrectanswer.\end{array}$