Answer:
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.
