Electricity

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AP Physics 1 › Electricity

Questions 1 - 10

Parallel circuit

What is the power in the below circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and ?

Explanation

To find the current we must first find the equivalent resistance. For resistors in parallel, the equivalent resistance is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...$

For this problem

$\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$

$\frac{1}{R_{eq}}=\frac{5}{30}$

$R_{eq}=\frac{30}{5}=6\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$

Now that we have found the current in the resistors we use the equation

to find the power in the circuit.

$P=IV=4.167*25\approx 104.167W$

Parallel circuit

What is the power in the below circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and ?

Explanation

To find the current we must first find the equivalent resistance. For resistors in parallel, the equivalent resistance is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...$

For this problem

$\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$

$\frac{1}{R_{eq}}=\frac{5}{30}$

$R_{eq}=\frac{30}{5}=6\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$

Now that we have found the current in the resistors we use the equation

to find the power in the circuit.

$P=IV=4.167*25\approx 104.167W$

Parallel circuit

What is the power in the below circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and ?

Explanation

To find the current we must first find the equivalent resistance. For resistors in parallel, the equivalent resistance is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...$

For this problem

$\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$

$\frac{1}{R_{eq}}=\frac{5}{30}$

$R_{eq}=\frac{30}{5}=6\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow$

$I=\frac{V}{R_{eq}}=\frac{25}{6}\approx 4.167A$

Now that we have found the current in the resistors we use the equation

to find the power in the circuit.

$P=IV=4.167*25\approx 104.167W$

What is the power of a circuit whose voltage is and equivalent resistance is $10\Omega$ ?

Explanation

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for current gives us $I=\frac{V}{R}$ .

Substituting this form of Ohm's law into the power equation gives us

$P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right$

The power equation is now in a form that we can solve with the information we are given.

$P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W$

What is the power of a circuit whose voltage is and equivalent resistance is $10\Omega$ ?

Explanation

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for current gives us $I=\frac{V}{R}$ .

Substituting this form of Ohm's law into the power equation gives us

$P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right$

The power equation is now in a form that we can solve with the information we are given.

$P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W$

What is the power of a circuit whose voltage is and equivalent resistance is $10\Omega$ ?

Explanation

The power in a circuit is determined by the equation $P=I^{2}R$ , where is the power of the circuit, is the current in the circuit, and is the equivalent resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, , where is the voltage, is the current in the circuit, and is the equivalent resistance of the circuit.

Solving Ohm's law for current gives us $I=\frac{V}{R}$ .

Substituting this form of Ohm's law into the power equation gives us

$P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right$

The power equation is now in a form that we can solve with the information we are given.

$P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W$

Light bulbs give their wattage based on their power output when they are in parallel with a voltage source. For most, that comes from an outlet which typically had a voltage of $\Delta V = 120V$ .

What is the resistance of a 60W lightbulb if it's plugged into a socket with $\Delta V = 120V$ ?

$240\Omega$

$720\Omega$

$\sqrt{780} \Omega$

$60\Omega$

Explanation

We can determine the resistance of the 60W lightbulb by using the equation that relates power, voltage, and resistance:

$\frac{\Delta V^2}{R}=P$

Where is resistance, $\Delta V$ is voltage difference in the circuit, and is the power output.

We know that the voltage difference is and that the power output is , so plug in and solve for the resistance:

$R=\frac{\Delta V^2}{P}=\frac{(120)^2V^2}{60J}=240\Omega$

Light bulbs give their wattage based on their power output when they are in parallel with a voltage source. For most, that comes from an outlet which typically had a voltage of $\Delta V = 120V$ .

What is the resistance of a 60W lightbulb if it's plugged into a socket with $\Delta V = 120V$ ?

$240\Omega$

$720\Omega$

$\sqrt{780} \Omega$

$60\Omega$

Explanation

We can determine the resistance of the 60W lightbulb by using the equation that relates power, voltage, and resistance:

$\frac{\Delta V^2}{R}=P$

Where is resistance, $\Delta V$ is voltage difference in the circuit, and is the power output.

We know that the voltage difference is and that the power output is , so plug in and solve for the resistance:

$R=\frac{\Delta V^2}{P}=\frac{(120)^2V^2}{60J}=240\Omega$

Light bulbs give their wattage based on their power output when they are in parallel with a voltage source. For most, that comes from an outlet which typically had a voltage of $\Delta V = 120V$ .

What is the resistance of a 60W lightbulb if it's plugged into a socket with $\Delta V = 120V$ ?

$240\Omega$

$720\Omega$

$\sqrt{780} \Omega$

$60\Omega$