Ohm’s Law, Electrical Power, Electrical Energy, Semiconductors, Superconductors Problems and Solutions

 Problem #1 

In Fig. 01a, a 20 Ω resistor is connected to a battery. Figure 01b shows the increase of thermal energy E_th in the resistor as a function of time t. The vertical scale is set by E_th,s = 2.50 mJ, and the horizontal scale is set by t_s = 4.0 s. What is the electric potential across the battery?

Fig.01

Answer;
Known:
Resistor, R = 20 Ω
E_th,s = 2.50 mJ = 2.50 x 10^─3 J
t_s = 4.0 s

First find the power absorbed by the resistor.

Remember that power is: Energy/time or  P = W/t 

Since this is a linear graph, we can pick any two arbitrary points and we'll get the same power which will be the slope in this case.

It's convenient to pick:  (0, 0) and (4, 2.50)

The power of the resistor is:

P = (2.50 x 10^─3 J ─ 0)/(4 s ─ 0) = 6.25 x 10^─4 Watts

The power in a resistor has many form such as:

P = Vi = I²R = V²/R.

The last one is the one that will suit us best:

6.25 x 10^─4 Watts = V²/20 Ω

 V² = 0.0125

V = 112 mV

Problem #2 
A certain brand of hot-dog cooker works by applying a potential difference of 120 V across opposite ends of a hot dog and allowing it to cook by means of the thermal energy produced. The current is 10.0 A, and the energy required to cook one hot dog is 60.0 kJ. If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously?

Answer;
Known:
Potential difference, V = 120 V
Current, i = 10.0 A
the energy required to cook one hot dog is W = 60.0 kJ = 60.0 x 10³ J

Electrical energy used to cook hot─dogs is given by

W = Vit

the time needed to cook a hot dog is given by

t = W/Vi = 3 x 60.0 x 10³ J/(120 V)(10.0 A) = 150 s

Problem #3
Thermal energy is produced in a resistor at a rate of 100 W when the current is 3.00 A. What is the resistance?

Answer;
Known:
Thermal energy is produced in a resistor at a rate is P = 100 W
the current is i = 3.00 A

The power in a resistor has many form such as:

P = Vi = I²R = V²/R.

V = P/i = 100 W/3.00 A = 33.33 V

Problem #4 
A 120 V potential difference is applied to a space heater whose resistance is 14 Ω when hot. (a) At what rate is electrical energy transferred to thermal energy? (b) What is the cost for 5.0 h at US$0.05/kWh?

Answer;
Known:
Potential difference is V = 120 V
resistance is R = 14 Ω

(a) electrical energy transferred to thermal energy is given by

P = Vi = (120 V)(14 Ω) = 1680 W

(b) the cost for 5.0 h at US$0.05/kWh is

Cost = 1680 W x 5.0 h x US$0.05/kWh = US$0.42

Problem #5
A heating element is made by maintaining a potential difference of 75.0 V across the length of a Nichrome wire that has a 2.60 x 10^─6 m² cross section. Nichrome has a resistivity of 5.00 x 10^─7 Ωm. (a) If the element dissipates 5000 W, what is its length? (b) If 100 V is used to obtain the same dissipation rate, what should the length be?

Answer;
Known:
Potential difference, V = 75.0 V
Resistivity, ρ = 5.00 x 10^─7 Ωm
cross-sectional area, A = 2.60 x 10^─6 m²

(a) If the element dissipates 5000 W, the length is determined by

P = V²/R

the resistance to a wire is determined by

R = ρL/A

Then, electric power is given by

P = V²A/ρL

5000 W = (75.0 V)²(2.60 x 10^─6 m²)/(5.00 x 10^─7 Ωm)L

L = 5.85 m

(b)  If 100 V is used to get the same level of dissipation, the length becomes

P = V²A/ρL

5000 W = (100 V)²(2.60 x 10^─6 m²)/(5.00 x 10^─7 Ωm)L

L = 10.4 m  

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