Question

Determine the magnitude of the current in the (a) 8.0- and (b) 2.0-Ω resistors in the drawing. Please include units. Thank you much.

http://oi67.tinypic.com/2e3vqet.jpg

Answer 1

Solution:

Applying Kirchhoff's current and voltage rules to solve the given circuit.

Let i be the current that comes out of 12 V battery. So, it flows towards the negative polarity of 4 V battery. But it gets split as i1 towards to 4V battery and i2 towards 2 ohm resistor.

So i = i1 + i2 by Kirchhoff's current law

Now by Kirchhoff's voltage law in case of loop having 4V bat, 8 ohm, 2 ohm resistor , we have algebraic sum of product of current and resistance i.e. 8 x i1 - 2 x i2 = 4   

Here clock wise is taken as positive by convention

Now for the lower loop with 2 ohm and 12 battery 2 x i2 = 12  

So, i2 = 6 A

Substituting this in second one, so we get, i1 = 2 A

From the first one i = 8 A

A) So in 8 ohm resistor, the current is 2 A from left to right

B) And in 2 ohm resistor the current is 6 A that too from left to right

Please rate my answer, good luck...