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高,那ol Physics : Electricity and Magnetism

Study concepts, example questions & explanations for High School Physics

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Example Questions

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Example Question #1 :Electricity And Magnetism

Calculate the magnitude of the electric field at a point that is located $\small 15cm$ directly north of a point charge, $\small Q = -15.0 * 10^{-6}C$ .

$k=9*10^9\frac{m^2N}{C^2}$

Possible Answers:

$3.0*10^5\frac{N}{C}$

$-6.0*10^6\frac{N}{C}$

$10.0*10^6\frac{N}{C}$

$-3.0*10^2\frac{N}{C}$

$-6.0*10^2\frac{N}{C}$

Correct answer:

$-6.0*10^6\frac{N}{C}$

Explanation:

The magnitude of an electric field due to a single point charge is given by the formula:

$E=k\frac{Q}{r^2}$

The variable $\small k$ refers to Coulomb's constant. We are given the charge and the distance between points. To use the equation, first convert the distance to meters.

$15cm*\frac{1m}{100cm}=0.15m$

We can now use the given constant, charge, and distance in the equation to solve for the magnetic field strength.

$E=(9*10^9\frac{m^2N}{C^2})(\frac{-15.0*10^{-6}C}{(0.15)^2})$

$E=(9*19^9\frac{m^2N}{C^2})(-6.67*10^{-4}\frac{C}{m^2})$

$E=-6.0*10^6\frac{N}{C}$

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Example Question #1 :Electric Circuits

What is the voltage of a circuit with 10A of current and $20\Omega$ of resistance?

Possible Answers:

200V

30V

10V

0.5V

Correct answer:

200V

Explanation:

For this problem, use Ohm's law: V=IR . In this equationis the voltage,is the current, andis the resistance.

Plug in the given values and solve for the voltage.

$V=10A* 20\Omega$

V=200V

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Example Question #1 :Circuit Calculations And Concepts

What is the current of a circuit with a voltage of 12V and a total resistance of $5\Omega$ ?

Possible Answers:

0.42A

60A

2.4A

4.8A

Correct answer:

2.4A

Explanation:

For this problem, use Ohm's law: V=IR . In this equationis the voltage,is the current, andis the resistance.

We can re-arrange the equation to solve specifically for.

$\frac{V}{R}=I$

Plug in the given values for voltage and resistance to solve for the current.

$\frac{V}{R}=I$

$\frac{12V}{5\Omega}=I$

2.4A=I

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问题# 3:例子Circuit Calculations And Concepts

What is the resistance in a circuit with a voltage of 50V and a current of?

Possible Answers:

$0.16\Omega$

$6.25\Omega$

$58\Omega$

$400\Omega$

$42\Omega$

Correct answer:

$6.25\Omega$

Explanation:

For this problem, use Ohm's law: V=IR . In this equationis the voltage,is the current, andis the resistance.

We can re-arrange the equation to solve specifically for.

$\frac{V}{I}=R$

Plug in the given values for voltage and current and solve for resistance.

$\frac{V}{I}=R$

$\frac{50V}{8A}=R$

$6.25\Omega=R$

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Example Question #4 :Circuit Calculations And Concepts

Abattery is connected to a circuit. The measured current is. What is the equivalent resistance?

Possible Answers:

$4.5\Omega$

$18\Omega$

$9\Omega$

$0.22\Omega$

$10\Omega$

Correct answer:

$4.5\Omega$

Explanation:

For this problem, use Ohm's law: V=IR .

We are given the voltage and current, allowing us to solve for the resistance.

9V=2A* R

$\frac{9V}{2A}=R$

$4.5\Omega=R$

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Example Question #1 :Electric Circuits

A closed electric circuit is set up so that there is a current of 9.9A and a voltage of 30.22V . What is the resistance in the circuit?

Possible Answers:

$299.18\Omega$

$20.32\Omega$

$6.1\Omega$

$3.05\Omega$

$0.33\Omega$

Correct answer:

$3.05\Omega$

Explanation:

For this problem, use Ohm's law: V=IR .

We are given the current and the voltage. Using these terms, we can solve for the resistance.

V=IR

30.22V=(9.9A) R

$\frac{30.22V}{9.9A}=R$

$3.05\Omega=R$

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Example Question #6 :Circuit Calculations And Concepts

The current in a circuit is 20A . If the voltage is 35V , what is the total resistance?

Possible Answers:

$0.57\Omega$

$3.5\Omega$

$700\Omega$

$1.75\Omega$

$70\Omega$

Correct answer:

$1.75\Omega$

Explanation:

For this problem use Ohm's law:

V=IR

We are given the current and the voltage, allowing us to solve for the resistance.

35V=(20A)*R

$R=\frac{35V}{20A}$

$R=1.75\Omega$

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Example Question #7 :Circuit Calculations And Concepts

An electrical circuit has a total of $3.22\Omega$ resistance and. What is the current?

Possible Answers:

0.089A

0.403A

11.22A

25.76A

2.48A

Correct answer:

2.48A

Explanation:

For this problem use Ohm's law:

V=IR

我们给出了电阻和电压,允许ing us to solve for the current.

$8V=I*(3.22\Omega)$

$I=\frac{8V}{3.22\Omega}$

I=2.48A

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Example Question #8 :Circuit Calculations And Concepts

An electrical circuit has a current ofand $32\Omega$ of resistance. What is the voltage?

Possible Answers:

0.25V

256V

40V

128V

Correct answer:

256V

Explanation:

For this problem use Ohm's law:

V=IR

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(8A)(32\Omega)$

V=256V

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Example Question #9 :Circuit Calculations And Concepts

An electrical circuit has a current of 13A and $26\Omega$ of resistance. What is the voltage?

Possible Answers:

13V

39V

338V

0.5V

Correct answer:

338V

Explanation:

For this problem use Ohm's law:

V=IR

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(13A)(26\Omega)$

V=338V

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高,那ol Physics : Electricity and Magnetism

Study concepts, example questions & explanations for High School Physics

所有高中物理资源

Example Questions

Example Question #1 :Electricity And Magnetism

Example Question #1 :Electric Circuits

Example Question #1 :Circuit Calculations And Concepts

问题# 3:例子Circuit Calculations And Concepts

Example Question #4 :Circuit Calculations And Concepts

Example Question #1 :Electric Circuits

Example Question #6 :Circuit Calculations And Concepts

Example Question #7 :Circuit Calculations And Concepts

Example Question #8 :Circuit Calculations And Concepts

Example Question #9 :Circuit Calculations And Concepts

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