I do one question first
We need to use kirchoff current's law
Let current through Vs = I1 (rightwards)
Let current through resistor R1 = I2 (downwards)
a) take one round around the left half
9 = (I1)(800) + (I2)(400)
800 I1 + 400 I2 = 9 -- (1)
Take the round around big rectangle
9 - 0.7 - 0.7 = (I1)(800) + (Id)(800)
800 I1 + 800 Id = 7.6 -- (2)
Finally, we know total current entering a junction = total
current exiting a junction, so
I1 = I2 + Id
Solve simultaneous equations of 3 variables
I1 = 8 mA
I2 = 6.5 mA
Id = 1.5 mA
Try repeating my method for part b
I look at the rest when got time
