Solving Circuit Problems: Finding Current in a Complex Circuit

[The Vin]

Homework Statement

The terminals A and B in Figure 21-31 are connected to a 9.0 V battery, where R1 = 1.8 and R2 = 3.2 .

Here is diagram of the circuit: http://www.webassign.net/walker/21-30alt.gif

I must find the current flowing through every resistor on the circuit.

Homework Equations

V=IR

The Attempt at a Solution

So I'm completely stuck on finding the current flowing through each resistor. I already calculated the total resistance of the entire circuit by breaking up the circuit and got a correct value of .914 ohms.

I know that for the 1.8 ohm resistor and the 2.5 ohm resistor, the voltage drop is from one terminal to the other, thus it's 9 V (the voltage of the battery) and the current flowing through those two resistors is 5 amps and 3.6 amps respectively via ohm's law.

But I'm not sure how to calculate the current for the 6.3 ohm resistor, 4.8 ohm resistor, 3.2 ohm resistor, or the 8.1 ohm resistor because I'm not quite sure how to determine the voltage drop across those resistors (I know it has to be less than 9 and that the voltage drop for 3.2 ohm resistor, the 4.8 ohm resistor, and the 8.1 ohm resistor will be the same since they are in parallel and then I can just subtract that voltage drop by 9 to get the Vdrop for the 6.3 ohm resistor. But I'm stuck in regards to finding this value.

Any help is appreciated.

 

[andrevdh]

Youve got the current through the first two resistors correct (R1 and 2.5 ohm).

For the rest of the circuit you could calculate the resultant resistance of the parallel combination (4.8, R2 and 8.1 ohm resistors). Then add this to the 6.3 ohm resistor (they are in series). Now you can calculate the current through this branch of the circuit (9 V over it since it acts as another parallel branch to the original two resistors, R1 and 2.5 ohm). This gives you the current through the 6.3 ohm resitor.

Then you can calculate the voltage drop over the 6.3 ohm resistor. The rest of the voltage (9 V minus this voltage drop) will appear over the parallel combination (R2, 4.8 and 8.1). Then you can calculate the current in each of these resisitors with this voltage.

 

[mjsd]

by the way, the total resistance I've got is 0.923 ohms

so now you know the total current coming out from the 9V source using V=IR. now this current I will be splitted into three portions: thru R1, thru the 2.5ohms and thru the branch containing the 3-parallel combinations plus the 6.3 ohms.

the faction of the total current going into each of this three branches will depend on the relative sizes(resistances). Now, observe that all these branches have the same potential difference across them namely V_AB or 9V. so again V=IR will tell you how the current will be splitted in three (given that you know the total resistance IN EACH BRANCH).
once you have found these 3 currents, all there is remaining is how current split within the 3-parellel combination. First you work out how the 9V is divided amongst the 6.3ohms and the 3-parallel combination. again V=IR to first find out voltage drop at the 6.3ohms then 9-(answer) gives you drop across the 3-parallel combination. after that you will have the voltage across the 3-parallel combination and you can split that current using method similar to the first splitting stage describe above.