How Does Resistance and EMF Affect Energy Transfer in a Circuit?

[justine411]

Homework Statement

A wire of resistance 5.0 Ω is connected to a battery whose emf ε is 2.0 V and whose internal resistance is 1.0 Ω. In 2.0 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

Homework Equations

P=iV
P=i^2r
P=iE(the fancy E)
Vb-Va=(E/(R+r))R
i=E/(R+r)

The Attempt at a Solution

I am having trouble finding the current, so I cannot find power.
I am using i=2.0V/(5.0+1.0)=0.333A
then I get Vb-Va=(0.333Ax5.0)
the answer for part a) is 80J
when I use P=iV I don't get that answer...

If I could make it more clear...I can, just post!

 

[justine411]

nevermind, I figured it out on my own

BOARD CLOSED

 

[m2003]

I would like to clarify the question and provide a more detailed response.

Firstly, it is important to understand the terms used in this question. Power is the rate at which energy is transferred or converted, measured in watts (W). Potential is the energy per unit charge, measured in volts (V). EMF (electromotive force) is the potential difference across a source, such as a battery, which causes current to flow.

Now, let's look at the given values in the question. The wire has a resistance of 5.0 Ω, the battery's EMF is 2.0 V, and its internal resistance is 1.0 Ω. We are asked to find the amount of energy transferred from chemical to electrical form in the battery, the amount of thermal energy dissipated in the wire, and the amount of thermal energy dissipated in the battery, all within a time frame of 2.0 minutes.

To find the current (i) in the circuit, we can use Ohm's Law: i = V/R, where V is the potential difference (in this case, the EMF of the battery) and R is the total resistance in the circuit (in this case, the sum of the wire's resistance and the battery's internal resistance). Thus, i = 2.0 V / (5.0 Ω + 1.0 Ω) = 0.333 A.

Now, to find the power (P) dissipated in the wire, we can use the formula P = i^2R, where i is the current and R is the resistance of the wire. Thus, P = (0.333 A)^2 x 5.0 Ω = 0.556 W. To find the energy dissipated in the wire over a period of 2.0 minutes, we can use the formula E = Pt, where P is the power and t is the time. Thus, E = 0.556 W x 120 s = 66.72 J.

To find the energy transferred from chemical to electrical form in the battery, we can use the formula E = εit, where ε is the EMF of the battery, i is the current, and t is the time. Thus, E = 2.0 V x 0.333 A x 120 s = 79.92 J.

Finally, to find the energy