What are the calculations for resistors in parallel?

[sskakam]

Homework Statement

In the circuit shown in the figure, the rate at which R1 is dissipating electrical energy is 20.0 W.

A) Find R1
B) Find R2
C) Find emf of battery
D) Find current through R2
E) Find current through 10 ohm resistor.
F) Find total electrical power consumption of the resistors
G) Find total electircal power delivered by battery

Homework Equations

Ohm's Law: V = IR
Resistors in parallel: I/V_ab = 1/R_eq = 1/R1 + 1/R2 + 1/R3
Current through resistors in parallel: I1 = V_ab/R1
Power delivered by battery: P = Ei
Energy dissipated by resistors: (i^2)R

The Attempt at a Solution

A) I set the power dissipation 20 = (i^2)R, having i = I1 = 2 A, and solved for R1.

B) I'm stuck here. Only thing I can think of using is the resistors in parallel equation but I don't know V, and I can't find V because I don't know R_eq yet.

C) Find R_eq and use V = IR_eq, which equals E.

D) I2 = V/R2

E) Same as part D

F) Apply the power consumption equation to each resistor and add them up.

G) P = Ei

 

[pooface]

Let us deal one step at a time.
A) good
B) There are three equations for power. P=VI, P=V^2/R, and P=I^2R. Think and execute.

Please show us some more of your work so that we can help if you need it.


<<post edited slightly by berkeman>>

 

[sskakam]

Many thanks for the quick reply.

B) 20 = V*I1, so V = 20/I1 = 10

I/V = 1/10 + 1/R1 + 1/R2
1/R2 = I/V - 1/10 - 1/5
R2 = (above)^-1 = 20

C) V = E in this case, right? E = 10

D) I2 = V/R2 = .5

E) (I'll call this one I3) I3 = V/10 = 1

F) (I3^2)10 + (I2^2)R2 +20 = 40

G) P = EI = 10*3.5 = 35? Can the power consumed by the resistors be greater than the power given by the battery? I think I made a mistake.

 

[pooface]

It should equal 35 Watts.
P3 = V*I =10V*1A=10W
P2 = V^2/R= 10V^2/20ohms=5W
P1 = 20W

 

[sskakam]

Ahh, thanks. I see what I did wrong. I forgot to square the I2 in (I2^2)R2.

Thank you very much.