oh sorry! i meant 7...the one where we have to use kirchoff's second law....i figured it out....current entering=current leaving...but if you could still have a look
yep its correct see if we take anticlockwise direction starting from E1
current is in the same direction so it will be
-I1R and then it faces the (+) terminal of E2 so we take it as negative so next is -E2
then we face the next resistor but here the current is moving in the opposite direction to our clockwise direction so it will be + I2R
then we face the next resistor, so the the current is the same direction to our chosen direction so it will be -I1R
and we then face the negative terminal of E1 so we take it as +E1
so we group them together
-I1R-E2+I2R-I1R+E1=0
got it?