I need help finding the current through all the resistors. I found the current through the 4ohm resistor is 2Amps. The current through the 6ohm resistor is 1.3 Amps but I got it wrong and have no idea how to get that number. I also have to find the currents and deltaVs for all the others. A start in the right direction would be much appreciated.

**Answer 1**

Start by working on the circuit as a whole. You have a 4.0 ohm resistor in series with a parallel track consisting of a 24 ohm resistor and a 6 ohm resistor in series with an 8 ohm and a 24 ohm resistor in parallel. It sounds a lot more complicated than it really is.

Start with the 8 and 24 in parallel. 1/r = 1/8 + 1/24 = 4/24 = 1/6, so the equivalent resistance of the 8 and 24 ohm resistor is 6.0 ohms. That’s in series with another 6.0 ohm resistor, making the total resistance of that leg 12.0 ohms.

So, we have a 24 ohm resistor in parallel with a 12 ohm resistor, giving us 1/r = 1/12 + 1/24 = 3/24 = 1/8, so the equivalent resistance of the parallel legs is 8.0 ohms.

This is in series with the 4.0 ohm resistor, giving you a total equivalent resistance of 12.0 ohms. Put it across a 24 V battery and I = V/R = 2.0 amps, as you had determined.

This, then, is the total current in the circuit. If you look at the diagram, you can see that all of the current in the circuit goes through the 4.0 ohm resistor, then splits to go through the two legs of the parallel branch. So, for the 4.0 ohm resistor, the voltage drop is V = IR = (2.0 A)(4.0 ohms) = 8.0 V.

This means that the other (24 – 8) = 16 volts is dropped across the parallel branch. Again, look at the diagram: the 24 ohm resistor is the only load in its branch, so it drops the whole 16 V. The current, then, is I = V/R = 16/24 = 2/3 A ~ 0.7 A

The remainder of the 2.0 A current, then, must go through the other branch. That means 2 – 2/3 = 4/3 A goes through there. All of that passes through the 6.0 ohm resistor, giving a voltage drop of V = IR = (4/3)(6) = 8.0 V.

That means that the remaining 8.0 V in that branch must be dropped by that inner parallel branch. 8 V dropped across the 8.0 ohm resistor puts 1.0 A of current through that. 8 V dropped across the 24 ohm resistor gives 1/3 A ~ 0.3 A current through that.

Check the totals and you’ll find all of them work.

**Answer 2**

Slightly different answers (I think your math was a little off) but thank you! the steps are all right!

**Answer 3**

1.[8×24]/[8+24]=6 ohms ….since they are in parallel

2.6+6=12..ohms………since they become series

3.[12×24]/[12+24]=8 ohms ….since they become parallel

4.4+8=12.ohms ……….since they become series

5.24/12=2A……….answer for circuit current

6.

a) PD & Current across 4 ohms = 2A x 4ohms= 8V & 2A

b)This Circuit current is divided though the TWO parallel branches which followed by

4 ohms resistor, a 24.ohms resistor in one branch & 12.ohms [equivalent] in branch consisting of resistors 6,8 &24 ohms in the first branch = 0.7A & 1.3 A……Circuit current 2A is divided into 1:2 ratio

as the 24 ohms will resist double times the 12 ohms resistor .

Now, Voltage across 24ohms =24V-8V( PD across 4 ohms)=16V (This is the remaining PD across the parallel branches 24 and [6+8parallel24] ohms .

16V is divided across [6+8parallel24] ohms (or) 6+6ohms[equivalent] into 1:1 ratio that is 8V & 8V

c)Now,the 1.3A is passing through the parallel paths of 8ohms & 24ohms in a ratio of 3:1,that is 0.975A : 0.325A