Question

If 2.0 × 10^-4 C of charge passes a point in 5.0 × 10^-6 s, what is the rate of current flow?

You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 Ω. If each light has a power rating of 0.50 W, what will happen?

Answer 1

This is not mathematics. But anyway:
\[I = \frac{dq}{dt} \rightarrow I =\frac{2*10^{-4}}{5*10^{-6}} = 0.4*10² = 40 C/s \ (Ampères) \]

Answer 2

Now for the second question...

Answer 3

Using Ohm's law, we have\[120 =(100*3)I,\ \ I = 0.4\ A\] Using the voltage divider rule, each lamp will have a voltage drop of 120/100 = 1.2 V.
Calculating the dissipated energy, we have P=RI²\[P = 1.2*0.4=0.48W\] The lamps will not burn out.

Answer 4

Oops, the dissipated energy in this line "Calculating the dissipated energy, we have P=RI²" should be P=VI. But it can be done like that too.

Answer 5

So for the first question, this; 4.0 × 10^-1 A
would be the answer?

Answer 6

No. 4.0 * 10 A, 40* 10^0 A or 0.4 * 10 A

Answer 7

I mean, 0.4 * 100 A

Answer 8

But the options I'm given are:
4.0 × 10^-1 A

4.0 × 10^1 A

1.0 × 10^-10 A

1.0 × 10^2 A

Answer 9

It's the second option! Remember that 10^1 is 10, 10^-1 = 1/10, 10^-2 = 1/10², and so on...

Answer 10

Ohhh, okay! Thanks so much