How Many Electrons Pass and What Are the Costs of Electrical Appliances?

[mustang]

Problem4.
A total charge of 12mC passes througha cross-sectional area of a nichrome wire in 3.1s.
How many electrons pass through the cross-sectional area in 30.0s? In units of electrons.
Note: What formula should I use?


Assume electrical energy cost $0.080 per k*W*h, and that appliances have a potential difference across them of 113 V.
Problem 16.
Calculate the cost of running an electric oven that draws 26.0A of current for 24hr. In units of $.
Note: How do I do this problem.

Problem 17.
Calculate the cost of running a television with a resistance of 27 ohms for 24hr. In units of $.

 

[fannemel]

Problem 4:
An electron has charge, and it's charge is the smallest amount of charge possible. Thus all greater charges equals a whole number of those.

Formula would be:
Nq = 12mC

where N = number of electron and q = e the electron charge = 1.60 * 10^-19

then you can find how many electrons that pass in 3.1 seconds and when you know that it's easy to find out how many that passes in 30s.

For the other problems use formulas
P = VI (P=W in your formula)
U = RI

 

[M98Ranger]

For Problem 4: To calculate the number of electrons passing through the cross-sectional area in 30.0s, we can use the formula Q = I*t, where Q is the total charge, I is the current, and t is the time. We know that Q = 12mC = 12*10^-3 C and t = 30.0s. We can rearrange the formula to solve for the number of electrons, which is given by n = Q/e, where e is the elementary charge. Plugging in the values, we get n = (12*10^-3 C)/(1.6*10^-19 C) = 7.5*10^16 electrons.

For Problem 16: To calculate the cost of running an electric oven, we can use the formula Cost = Power * Time * Cost per kWh. We know that the power, P, is given by P = V*I, where V is the potential difference and I is the current. Plugging in the values, we get P = (113 V)*(26.0 A) = 2938 W. The time, t, is given as 24 hours. To convert this to kWh, we divide by 1000, so t = 24 hr * (1kWh/1000 W) = 24 kWh. The cost per kWh is given as $0.080. Plugging all these values into the formula, we get Cost = (2938 W)*(24 kWh)*($0.080/kWh) = $56.22.

For Problem 17: To calculate the cost of running a television, we can use the same formula as in Problem 16. The only difference is that we need to calculate the power, P, using Ohm's Law, which states that V = I*R, where V is the potential difference, I is the current, and R is the resistance. Plugging in the values, we get P = (113 V)*(27 ohms) = 3051 W. The rest of the calculation follows the same steps as in Problem 16, and the cost is found to be $58.44.