Question: Consider the circuit ( Assume R 1 = 1 2 . 0 Omega , R 2 = 2 . 2 0 Omega

 Consider the circuit  (Assume R1=12.0\Omega , R2=2.20\Omega , and V =7.90 V.)

A rectangular circuit begins at the positive terminal of a battery labeled V, which is on the bottom side of the rectangle. The circuit extends up and to the left to resistor labeled R2 on the top side of the rectangle. To the right of R2 the circuit splits into two parallel horizontal branches.
The top branch splits into two parallel horizontal sub-branches. There is a resistor labeled R1 on the top sub-branch and a 5.00 resistor on the bottom sub-branch. The sub-branches then recombine and the top branch extends to the right to a 4.00 resistor.
The bottom branch contains a 3.00 resistor.
The top and bottom branches then recombine and the circuit extends down and to the left until it reaches the negative terminal of the battery.
(a) Calculate the equivalent resistance of the R1 and 5.00-\Omega resistors connected in parallel.
\Omega

(b) Using the result of part (a), calculate the combined resistance of the R1,5.00-\Omega and 4.00-\Omega resistors.
\Omega

(c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00-\Omega resistor.
\Omega

(d) Combine the equivalent resistance found in part (c) with the R2 resistor.
\Omega

(e) Calculate the total current in the circuit.
A

(f) What is the voltage drop across the R2 resistor?
V

(g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00-\Omega resistor.
V

(h) Calculate the current in the 3.00-\Omega resistor.
A

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