Solving DC Circuits: Questions 23 & 35

[physpopib]

Been stumped on these 2 questions for at least a week.
Both diagrams scanned + uploaded.
23.Find the current in the 8 ohm resistor.
1/Req1=1/8+1/4
2.67+6=8.67
1/Req2=1/8.67+1/10
R=4.27+.5+5=10.14
I first found the voltage drop in each resistor.
I=V/R
I=9/10.14=.89A
volt.drop .5 resistor=.5*.89=.445V
volt.drop 5 resistor=5*.89=4.45V
volt.drop 4.27 resistor=4.27*.89=3.8V
Then I solved for current in the 8 ohm resistor.
I=[9V-(.89A)*(.5*5)]/x ohm
And that's where I'm stuck. Don't know what ohm to divide by. I think 25 but I don't know why.
35.What would current I1 be if the 12 ohm resistor were shorted out(r=1 ohm)?
OK...this is where I'm just screwed by Kirchhoff's Laws.
I'm thinking:
I3=I2+I1
Loop 1: 12=(1)I1+(8)I2
Loop 2: 12=(1)I2+(10)I2+(12)I2
Loop 3: 6=(1)I3+(18)I3+(15)I3+(10)I2+(1)I2
Completely stuck. Help please.

 

[Davorak]

Pro 23.

"I=[9V-(.89A)*(.5*5)]/x ohm"
No quite.
Use Req2 to find the voltage drop over that system of resistors. Or
VReq2=[9V-(.89A)*(.5*5)]

Then find the current going through Req1 + 6. Now you can find the voltage drop over Req1. Once you have the voltage drop you can find the current.

Does this make sense?

 

[thepatientmental]

For question 23, it looks like you have all the right steps in finding the current in the 8 ohm resistor. However, instead of trying to solve for the unknown resistance (x), you can simply use the current you found in the 10.14 ohm equivalent resistor (0.89A) and apply Ohm's Law to find the current in the 8 ohm resistor.

I = V/R = 9V/8 ohm = 1.125A

For question 35, you are on the right track with using Kirchhoff's Laws. However, you need to take into account that when the 12 ohm resistor is shorted out, it no longer has any resistance and can be ignored in the circuit. This means that the current in loop 1 (I1) will now only flow through the 1 ohm resistor and the 8 ohm resistor.

Using Kirchhoff's Laws, you should get the following equations:

Loop 1: 12 = (1)I1 + (8)I2
Loop 2: 12 = (1)I2 + (10)I2 + (12)I2

Solving for I1, you will get:

I1 = (12 - (8)I2)/1

Substituting this into Loop 2, you will get:

12 = (1)I2 + (10)I2 + (12)I2 + (12 - (8)I2)
12 = (15)I2
I2 = 0.8A

Now, you can use this value of I2 to solve for I1:

I1 = (12 - (8)(0.8))/1 = 5.6A

So, when the 12 ohm resistor is shorted out, the current in the 1 ohm resistor (I1) will be 5.6A.

I hope this helps and clarifies the steps you need to take in solving these two problems. Remember to always double check your equations and units when solving DC circuits. Good luck!