Question

In the circuit diagram given below,suppose resistors R1,R2 and R3 have the values 4,6 and 8 ohms respectively ,which have been connected to a battery of 12 V.calculate
1》the current through each resistor.
2》the total current in the circuit
3》the total circuit resistance

Answer 1

1) Current passing through R1:

V = IR

I = 12/4= 3A

Current passing through R2:

V = IR

I = 12/6= 2A

Current passing through R3:

V = IR

I = 12/8= 3/2A

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You haven't mentioned whether the resistors are connected in parallel or series. But in both you have to find Req. and then using V = IR find the current for the 2nd question and through finding  req. u get the answer for your 3rd question

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Answer 2

For finding current through each resistor

Apply the formula

R=v/I

Therefore

12/4 =3A

12/6=2A

12/8=1.5A

This was the solution for the question A

NOW FOR QUESTION B

Total current

Therefore

I= I1+I2+I3

=3+2+1.5

=6.5A

This was the solution for question B

Now for question c

If the circuit is series circuit

Then

R=R1+R2+R3

=4+6+8=18ohm

This is not possible because resistance cannot be more than current

Therefore the connection is in the parallel circuit

So, according to the formula

1/R=1/R1+1/R2+1/R3

=1/4+1/6+1/8

=6+4+3/24

1/R=13/24

Therefore

R=24/13=1.85 ohm is the total resistance of the circuit