Voltage, Current, and Resistance Calculations

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a1234
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Homework Statement



I'm asked to find the voltage, current, and resistance through each resistor in the given circuit.

Homework Equations



Req for parallel = 1/R1 + 1/R2 + ...
Req for series = R1 + R2 + ...

The Attempt at a Solution



First, I want to find the total resistance in the circuit.
Since R3 and R5 are in series, is R35 = 18 ohms?
Then R4 and R35 would be in parallel, so 1/9 + 1/18 = 1/Req
Req = 6 ohms

6 + 9 + 2.5 = 17.5 ohms
Is this the total resistance in the circuit?
 

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a1234 said:

Homework Statement



I'm asked to find the voltage, current, and resistance through each resistor in the given circuit.

Homework Equations



Req for parallel = 1/R1 + 1/R2 + ...
Req for series = R1 + R2 + ...

The Attempt at a Solution



First, I want to find the total resistance in the circuit.
Since R3 and R5 are in series, is R35 = 18 ohms?
Then R4 and R35 would be in parallel, so 1/9 + 1/18 = 1/Req
Req = 6 ohms

6 + 9 + 2.5 = 17.5 ohms
Is this the total resistance in the circuit?
Yes, that is all correct.
 
Here are the calculations I have so far:
R1: 0.69 amps, 6.17 volts
R2: 0.69 amps, 1.17 volts
R4: 1.33 amps, 12 volts

For R3 and R5, am I supposed to consider them to be in series with the battery, or in parallel?
 
a1234 said:
Here are the calculations I have so far:
R1: 0.69 amps, 6.17 volts
R2: 0.69 amps, 1.17 volts
R4: 1.33 amps, 12 volts

For R3 and R5, am I supposed to consider them to be in series with the battery, or in parallel?
I agree with the current through R1 and R2 (but closer to 0.68?), and the voltage across R1.
Your voltage across R2 looks like a typo (transposed digits).
There is no way that R4 can have a larger current than R1 and R2.
 
For R2, the voltage is 0.69 * 2.5 = 1.725. I don't know how I ended up with 1.17.
For R4, I thought it was in parallel and did 12 = 9 * I to get I = 1.33 amps.
 
a1234 said:
For R2, the voltage is 0.69 * 2.5 = 1.725. I don't know how I ended up with 1.17.
For R4, I thought it was in parallel and did 12 = 9 * I to get I = 1.33 amps.
The current that goes through R1 gets split, some going through R4, the rest through R5 and R3.
 
Would the voltage going through R4 be different from 12 V?
 
a1234 said:
Would the voltage going through R4 be different from 12 V?
The voltage across R4 will be less than 12V. What is the voltage drop across R1?
 
12/17.5 = 0.69
0.69 * 9 = 6.17 volts
 
a1234 said:
12/17.5 = 0.69
0.69 * 9 = 6.17 volts
Right, and you also calculated the drop across R2. So what is the drop across R4?
 
Would we work backwards to get 12 - 6.17 - 1.17 = 4.66, and divide that by 3?
 
a1234 said:
Would we work backwards to get 12 - 6.17 - 1.17
1.71
a1234 said:
and divide that by 3?
why 3?
 
The voltage drop across R4 is 12 - 6.17 - 1.71 = 4.12 V.
 
a1234 said:
The voltage drop across R4 is 12 - 6.17 - 1.71 = 4.12 V.
Right, so what is the current through it?