Problem #1
In Fig. 01, the ideal batteries have emfs ε1 = 12 V and ε2 = 6V. What are (a) the current, the dissipation rate in (b) resistor 1 (4.0 Ω) and (c) resistor 2 (8.0 Ω), and the energy transfer rate in (d) battery 1 and (e) battery 2? Is energy being supplied or absorbed by (f) battery 1 and (g) battery 2? |
| Fig.01 |
Known:
emfs ε1 = 12 V
emfs ε2 = 6 V
resistor 1, R1 = 4.0 Ω
resistor 1, R2 = 8.0 Ω
(a) the current
First solve for the current in the circuit:
ε1 − iR2 − ε2 − iR1 = 0
i = (ε1 – ε1)/(R1 + R2)
i = (12 V – 6 V)/(4.0 Ω + 4.0 Ω) = 0.50 A
(b) the dissipation rate in resistor R1
P1 = i2R12 = (0.50 A)2(4.0 Ω) = 1.0 W
(c) the dissipation rate in resistor R2
P2 = i2R22 = (0.50 A)2(8.0 Ω) = 2.0 W
(d) the energy transfer rate in battery 1
P1 = ε1i = 12 V x 0.50 A = 6.0 W
(d) the energy transfer rate in battery 2
P2 = ε2i = 6 V x 0.50 A = 3.0 W
(f) Battery 1 supplies energy
(g) Battery 2 absorbs energy
Problem #2
In Fig. 02, the circuit shown, the ideal batteries have EMFs of ε1 = 150 V and ε2 = 50 V and the resistances are R1 = 3.0 Ω and R2 = 2.0 Ω. If the potential at P is defined to be 100 V, what is the potential at Q?
 |
| Fig.02 |
Answer;
Known:
emfs ε1 = 150 V
emfs ε2 = 50 V
the resistances are R1 = 3.0 Ω
the resistances are R2 = 2.0 Ω
First solve for the current in the circuit:
ε1 − iR2 − ε2 − iR1 = 0
i = (ε1 – ε1)/(R1 + R2)
i = (150 V – 50 V)/(3.0 Ω + 2.0 Ω) = 20 A
Now let’s add the potential difference to go from point P to Q:
VQ = VP – ε1 + iR2
VQ = 100 V – 150 V + (20 A)(2.0 Ω) = –10 V
Problem #3
A car battery with a 12 V emf and an internal resistance of 0.040 Ω is being charged with a current of 50 A. What are (a) the potential difference V across the terminals, (b) the rate Pr of energy dissipation inside the battery, and (c) the rate Pemf of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) V and (e) Pr?
Answer:
Known:
emfs ε = 12 V
internal resistance, r = 0.040 Ω
current, I = 50 A
(a) the potential difference V across the terminals is
V = ε + ir = 12 V + 50 A x 0.040 Ω = 14 V
(b) the rate Pr of energy dissipation inside the battery is
Pr = i2r = (50 A)2 x 0.040 Ω = 100 W
(c) the rate Pemf of energy conversion to chemical form is
Pemf = εi = 12 V x 50 A = 600 W
(d) When the battery is used to supply 50 A to the starter motor, then V is
V = ε ─ ir = 12 V ─ 50 A x 0.040 Ω = 10 V
(e) When the battery is used to supply 50 A to the starter motor, then Pr is
Pr = i2r = (50 A)2 x 0.040 Ω = 100 W
Problem #4
A standard flashlight battery can deliver about 2.0 Wh of energy before it runs down. (a) If a battery costs US$0.80, what is the cost of operating a 100 W lamp for 8.0 h using batteries? (b) What is the cost if energy is provided at the rate of US$0.06 per kilowatt-hour?
Answer:
Known:
energy delivered by on battery is 2.0 W.h
(a) If the battery costs US$ 0.80, then the cost of operating a 100 W lamp for 8.0 hours is
100 x 8/2 x US$ 0.80 = US$ 320
(b) the cost if energy is provided at the rate of US$0.06 per kilowatt-hour.
The total energy consumed by the lamp is 800 Wh = 0.8 kWh
So, the cost is 0.8 kWh x US$0.06 = US$0.048
Problem #5
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 form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?
Answer:
Known:
wire of resistance, R = 5.0 Ω
emf, ε = 2.0 V
internal resistance, r = 1.0 Ω
(a) Total energy that is transferred from chemical to electrical form in 𝑡 = 4 min
𝑈𝑡𝑜𝑡𝑎𝑙 = ℇ𝐼t
Where
i = ε/(R + r) = 2.0 V/(5.0 Ω + 1.0 Ω) = 1/3 A
So, 𝑈𝑡𝑜𝑡𝑎𝑙 = ℇ𝐼r = 2.0 V x 1/3 A x 2 x 60 s = 80 J
(b) dissipated as thermal energy in the wire
𝑈wire = i2Rt = (1/3 A)2(5 Ω)(2 x 60 s) = 67 J
(c) dissipated as thermal energy in the battery is
𝑈thermal = 𝑈𝑡𝑜𝑡𝑎𝑙 ─ 𝑈wire = 80 J ─ 67 J = 13 J
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