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1. The diagram attached shows part of an electric circuit. What is the current through resistor R_{1}?
To summarize:
I=6.0 A
R_{1} = 11.0 Ω
R_{2} = 33.0 Ω
R_{3} = 67.0 Ω
R_{2} and R_{3} are in parallel, and R_{1} and R_{2 and 3} are in parallel.
I know we aren't supposed to attach pictures of questions, but a diagram does make my question a lot easier to understand.
V=IR
I_{T in series} = I_{1}=I_{2}=I_{3}
I_{T in parallel} = I_{1}+I_{2}+I_{3}
R_{T in series} = R_{1}+R_{2}+R_{3}
R_{T in parallel} = [(1/R_{1})+(1/R_{2})+(1/R_{3})]^{1}
Using the resistor laws, I can find resistance of R_{2 and 3} by (1/33+1/67)^{1}.
Using Kirchoff's Laws, I know that the resistance coming out has to equal the 6.0 A.
But I don't get how all this information is relevant. The question doesn't give me voltage or any other information and I don't know how to figure out how much current goes through each resistor.
To summarize:
I=6.0 A
R_{1} = 11.0 Ω
R_{2} = 33.0 Ω
R_{3} = 67.0 Ω
R_{2} and R_{3} are in parallel, and R_{1} and R_{2 and 3} are in parallel.
I know we aren't supposed to attach pictures of questions, but a diagram does make my question a lot easier to understand.
Homework Equations
V=IR
I_{T in series} = I_{1}=I_{2}=I_{3}
I_{T in parallel} = I_{1}+I_{2}+I_{3}
R_{T in series} = R_{1}+R_{2}+R_{3}
R_{T in parallel} = [(1/R_{1})+(1/R_{2})+(1/R_{3})]^{1}
The Attempt at a Solution
Using the resistor laws, I can find resistance of R_{2 and 3} by (1/33+1/67)^{1}.
Using Kirchoff's Laws, I know that the resistance coming out has to equal the 6.0 A.
But I don't get how all this information is relevant. The question doesn't give me voltage or any other information and I don't know how to figure out how much current goes through each resistor.
Attachments

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