PART A: Determine the magnitude of the current that flows through 6.0 -V battery

Question

PART A: Determine the magnitude of the current that flows through 6.0 -V battery when the switch is closed?

PART B: Determine the magnitude of the current that flows through 9.0 -V battery when the switch is closed?

Solution: (a) and (b): [When switch closed]

Applying KVL for left loop:

2 i1 - 4 i2 = 6 - 9 2 i1 - 4 i2 = -3 ...(1)

Applying KVL for right loop:

4 i2 + 5 ( i1 + i2 ) = 9 5 i1 + 9 i2 = 9 ...(2)

I know that answers will be

i1 = 0.237 A

i2 = 0.868 A

I understand the phsyics behind the problem, but I'm just having trouble with the algebra and figuring out the equations to get the currents I'm looking for. Thanks in advance!

Explanation / Answer

lets assume i1 is in the upward direction in battery 6v

and i2 is in the battery 9v in the upward direction

so we have i1+12 in 5ohm in downward direction on closing the switch

now here we are using 2 loops:

ist one is of comprised of 6v battery 2 ohm and 4 ohm and 9 volt battery

2nd one is the 9v battery 4ohm and 5 ohm resistance

now equations that you have are:

2i1-4i2 =-3 ----- 1

4 i2 + 5 ( i1 + i2 ) = 9

=>5 i1 + 9 i2 = 9 -----2

now find value of i1 in terms of i2 from eq1 and put in eq 2 i.e

i1= (4i2 - 3 )/2

putting it in eq 2

5 ( (4i2 - 3 ) /2 ) + 9i2 = 9

=> 20 i2 -15 +18i2 = 18

=>38 i2 =33

=> i2= 33/38 =0.868

so from i2 we calculate i1 = (4x 0.868 - 3) / 2 = 0.237 A

plz comment for any questions or doubt help thanks!!

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