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AP Physics C Electricity : Calculating Current

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Example Questions

Example Question #1 : Current

A lamp has a 60W $\displaystyle 60W$ bulb. If the house wiring provides 110V $\displaystyle 110V$ to light up that bulb, how much current does the bulb draw?

Possible Answers:

$0.72\ \text{A}$ $\displaystyle 0.72\ \text{A}$

$0.55\ \text{A}$ $\displaystyle 0.55\ \text{A}$

$1.37\ \text{A}$ $\displaystyle 1.37\ \text{A}$

$5.83\ \text{A}$ $\displaystyle 5.83\ \text{A}$

$0.03\ \text{A}$ $\displaystyle 0.03\ \text{A}$

Correct answer:

$0.55\ \text{A}$ $\displaystyle 0.55\ \text{A}$

Explanation:

The formula for power is , P=IV $\displaystyle P=IV$and we are given the following values.

$P=60\ \text{W}$ $\displaystyle P=60\ \text{W}$

$V=110\ \text{V}$ $\displaystyle V=110\ \text{V}$

Solve for the current, $\displaystyle I$.

$I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$ $\displaystyle I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$

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Example Question #13 : Electric Circuits

A particle accelerator with a radius of 500 meters can have up to $10^{15}$ $\displaystyle 10^{15}$ protons circulating within it at once.

$q_{proton}=1.60*10^{-19}C$ $\displaystyle q_{proton}=1.60*10^{-19}C$

How fast must the protons in the accelerator move in order to produce a current of 1A?

Possible Answers:

$9.21*10^{7}\frac{m}{s}$ $\displaystyle 9.21*10^{7}\frac{m}{s}$

$3.00*10^8\frac{m}{s}$ $\displaystyle 3.00*10^8\frac{m}{s}$

$2.04*10^8\frac{m}{s}$ $\displaystyle 2.04*10^8\frac{m}{s}$

$1.95*10^6\frac{m}{s}$ $\displaystyle 1.95*10^6\frac{m}{s}$

$1.96*10^7\frac{m}{s}$ $\displaystyle 1.96*10^7\frac{m}{s}$

Correct answer:

$1.96*10^7\frac{m}{s}$ $\displaystyle 1.96*10^7\frac{m}{s}$

Explanation:

The current produced is the total charge that circulates the particle accelerator per unit time.

$I=\frac{dQ}{dt}$ $\displaystyle I=\frac{dQ}{dt}$

We calculate this by the equation:

$I = \frac{nqv}{2\pi R}$ $\displaystyle I = \frac{nqv}{2\pi R}$

$\displaystyle n$ is the number of protons, $\displaystyle q$ is the charge per proton, $\displaystyle v$ is the velocity of each proton, and $\displaystyle R$ is the radius of the particle accelerator.

Using the given current, we then solve for the velocity:

$1A = \frac{(10^{15})(1.60*10^{-19}C)v}{2\pi (500m)}$ $\displaystyle 1A = \frac{(10^{15})(1.60*10^{-19}C)v}{2\pi (500m)}$

$v=\frac{(1A)2\pi(500m)}{(10^{15})(1.60*10^{-19}C)}= 1.96*10^7\frac{m}{s}$ $\displaystyle v=\frac{(1A)2\pi(500m)}{(10^{15})(1.60*10^{-19}C)}= 1.96*10^7\frac{m}{s}$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Calculating Current

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Current

Example Question #13 : Electric Circuits

All AP Physics C Electricity Resources

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