Question about Resistors in Physics?

Question

A 56.0 W resistor is connected in parallel with a 118 W resistor. This parallel group is connected in series with a 15.0 W resistor. The total combination is connected across a 18.0 V battery. (a) Find the current and (b) the power disspated in the 118- W resister.

Answer 1

First calculate the resistance in parallel.

1/Rt = 1/R1 + 1/R2

1/Rt = 1/56 + 1/118

Then calculate the series by taking the answer from above and adding 15.

Once you have the total R for the parallel-series circuit you calculate the current using

V = IR ; 18 V = I (Rt from above)

Answer 2

First you need to simply the resistors. Combine the 2 parallel using the rule **Req means equilvant resistor**:
1/Req= 1/R1+ 1/R2 (for parallel)
therefore you have: 1/Req = (1/56)+(1/118).
do the math and you have Req=37.98 approx 38

since you combined them, now the whole system is a series circuit. which can be combine: Req = R1+R2 (for series)
Req= 38+15 = 53.

(a)
In order to find the current use Ohm's law: V=IR
so we have the voltage and resistor, therefore: I = V/R
I = 18/53 = .0339 aprrox 0.34 amps

(b) for power use the formula: P= I^2*R
P= (0.34)^2* 118 = 13.64 W