What is the current in a circuit with multiple resistors and a given voltage?

[cjc881]

Homework Statement

In the circuit below find the current I3 (in A) when R1 = 26 Ω, R2 = 52 Ω, R3 = 36 Ω, R4= 25 Ω, R5 = 162 Ω, and V = 80 V.
Be careful about the sign.

Homework Equations

I=V/R

The Attempt at a Solution

I tried to solve for Req...i don't think I'm doing it right though. I calculated R2 being in parallel with R4. Then I said that answer was in series with R1 and R3 (no idea if that's right). Then i said that answer was in parallel with R5 and got Req to equal 53.05Ω

 

[gneill]

Homework Statement

In the circuit below find the current I3 (in A) when R1 = 26 Ω, R2 = 52 Ω, R3 = 36 Ω, R4= 25 Ω, R5 = 162 Ω, and V = 80 V.
Be careful about the sign.

Homework Equations

I=V/R

The Attempt at a Solution

I tried to solve for Req...i don't think I'm doing it right though. I calculated R2 being in parallel with R4. Then I said that answer was in series with R1 and R3 (no idea if that's right). Then i said that answer was in parallel with R5 and got Req to equal 53.05Ω

Hi cjc881, Welcome to Physics Forums.

R2 is indeed in parallel with R4, but if you look closely at the circuit you will see that the resulting resistance is not in series with R3. It often helps if you redraw the circuit with the "reduced" components in place so that you can more easily see what the next reduction opportunity might be.

In the above version of the circuit R24 is the resistance formed by the parallel combination R2 || R4. Can you see a new opportunity to combine resistors?

 

[cjc881]

Is R3 and R24 in parallel with R5?

 

[gneill]

Is R3 and R24 in parallel with R5?

No. In order to be in parallel two components must share exactly two nodes. As you can see, R3 does not share two nodes with R5, and neither does R24 share two nodes with R5. However...

For two components to be in series they must be connected at a node that is not shared with any other connections. Can you spot such a pair of resistors in the "new" layout?

 

[cjc881]

R24 in series with R5?

 

[NascentOxygen]

R24 in series with R5?

yes it is

 

[cjc881]

so then would R3 be in parallel with the resulting R245?
and then that answer be in series with R1?

 

[NascentOxygen]

so then would R3 be in parallel with the resulting R245?
and then that answer be in series with R1?

 

[cjc881]

So to find the current I3 would I:
find the circuit current by dividing V/Req (80V/55.97Ω)

say R1 will have that value passing through it (1.429 A)

calculate the v drop across R1 = (26Ω x 1.429A)

Subtract 80v by 37.16 V

Then say R3 will have (42.84V/36Ω)=1.19A running through it?

 

[NascentOxygen]

That looks right.

 

[cjc881]

sweet! thanks for the help!