A wire of resistance 5.0 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 0.61 Ω. In 2.9 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?

a. 496.26J

b. 442.29J

c. 53.96J

Explanation:

Given that,

Resistance of wire is 5.0 Ω

R = 5.0 Ω

Connected to EMF of 4V

ε = 4V.

Internal resistance is 0.61Ω

Time t = 2.9mins

t = 2.9 * 60

t = 174 seconds

Power dissipated in the circuit

The equivalent resistance of a series connection is

Req = R1 + R2

Req = R + r

Req = 5+0.61

Req = 5.61Ω

The power dissipation is given as

P = IV

From ohms law V=iR

Then, I=V/R

So, P = (V/R) * V

P = V²/R

P = 4² / 5.61

P = 16/5.61

P = 2.852 Watts

a. The energy transfer from chemical to the electrical is give as

Power = Energy / Time

P = E/t

E = Pt

E = 2.852 * 174

E = 496.26J

The energy transfer from chemical to the electrical as form of heat is 496.26J

b. Energy dissipated in the wire

The resistance of the wire is 5Ω

The current flowing in the circuit is give as

V = I (r + R)

4 = I (0.61 + 5)

4 = 5.61 * I

Then, I = 4/5.61

I = 0.713 Amps

Then the current flowing in the circuit is 0.713A

So, the energy dissipated it the wire of 5Ω is give as

E = IVt

Where V = IR from ohms law

E = I²Rt

E = 0.713² * 5 * 174

E = 442.29 J

The energy dissipated in the wire as form of heat is 442.29J.

c. The energy dissipated in the battery

Since, the internal resistance of the battery is 0.61Ω

Then, energy dissipated is given as

E = I² Rt

E = 0.713² * 0.61 * 174

E = 53.96J

The energy dissipated in the battery it to its internal resistance as form of heat is 53.96J.