JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Current Electricity

Label the two mesh currents and apply Kirchhoff's voltage law. Around the loop carrying the $20\text{V}$ source, $-20+2i_1+10(i_1+i_2)+5i_1=0$, that is $17i_1+10i_2=20$. Around the loop carrying the $10\text{V}$ source, $-10+4i_2+10(i_1+i_2)=0$, that is $10i_1+14i_2=10$. Solving the pair gives $i_2\approx -0.21\text{A}$; the negative sign simply says the current in the $10\text{V}$ battery runs from its positive to its negative terminal, with a magnitude close to $0.21\text{A}$.