Ohm's Law - AP Physics 2

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Question

$R_1=10\Omega$

$R_2=20\Omega$

What is the current through the battery in the above circuit?

Answer

First, find the total resistance of the circuit. Since the resistors are in parallel, use the following formula:

$\frac{1}{R1}+\frac{1}{R2}=\frac{1}{R_{total}}$

Plug in known values.

$\frac{1}{10\Omega}+\frac{1}{20\Omega}=\frac{1}{R_{total}}$

$R_{total}=6.66\Omega$

Next, use Ohm's law to find current.

Plug in known values.

$I=\frac{25V}{6.66\Omega}=3.75A$

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Question

You have a circuit with a $5\Omega$ resistor connected to a battery. What is the current through the resistor?

Answer

To find the current in a circuit with a battery and resistor(s), you use Ohm's Law.

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

We have the voltage and we have the resistance, so we don't need to rearrange the equation.

$I=\frac{10}{5}$

Therefore, the current through the resistor is 2 amps ().

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Question

There are 3 resistors in series. Their resistances are, in order, $2\Omega$ , $4\Omega$ , and $6\Omega$ . The total potential drop is . What is the potential drop across the second resistor?

Answer

Use Ohm's law to find the current passing through each resistor. Because they are in series, they have the same amount of current. Once we get the current, we can plug in the resistance for each resistor to find its potential drop.

$I&=\frac{V_{tot}}{R_{tot}}$

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

Now, find the potential drop across the $4\Omega$ resistor.

$V=(1A)(4\Omega)$

Therefore, the potential drop across the $4\Omega$ resistor is

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Question

A battery produces a current of in a piece of copper wire. What is the resistance of the copper wire?

Answer

Even though there is no resistor, Ohm's law still applies. Use it to find the resistance of the wire.

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

$R= \frac{4.5V}{0.90A}$

$R= 5\Omega$

The resistance of the copper wire is $5\Omega$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the voltage drop across .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find the equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R_{1+2}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current through the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

Since and are in parallel, they will have the same voltage drop accross them.

$V_{1+2}=\frac{9}{8}A \cdot R_{1+2}$

$V_{1+2}=\frac{9}{8}A\cdot 2\Omega$

$V_{1+2}=\frac{9}{4}V$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the voltage drop across .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

and will have the same voltage drop across them, as they are in parallel, and are equivalent to the combined resistor $R_{1+2}$

$V=\frac{9}{8}A\cdot R_{1+2}$

$V=\frac{9}{8}A\cdot 2\Omega$

$V=\frac{9}{4}V$

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Question

$R_1=R_3=R4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the voltage drop across .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

and will have the same voltage drop across them, as they are in parallel, and are equivalent to the combined resistor $R_{1+2}$

$V=\frac{9}{8}A\cdot R_{3}$

$V=\frac{9}{8}A\cdot 3\Omega$

$V=\frac{27}{8}V$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the current through .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

The current through and needs to add up to the total current, since they are in parallel.

$I_1 + I_2 =I_{total}$

$I_1 + I_2 =\frac{9}{8}A$

Also, the voltage drop across them need to be equal, since they are in parallel.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

Set up a system of equations.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

$I_1 + I_2 =\frac{9}{8}A$

Solve.

$1.5I_1=\frac{9}{8}A$

$I_1=\frac{3}{4}A$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the current through .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

The current through and needs to add up to the total current, since they are in parallel.

$I_1 + I_2 =I_{total}$

$I_1 + I_2 =\frac{9}{8}A$

Also, the voltage drop across them need to be equal, since they are in parallel.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

Set up a system of equations.

$I_1\cdot3\Omega=I_2\cdot 6\Omega$

$I_1 + I_2 =\frac{9}{8}A$

Solve.

$3I_2=\frac{9}{8}A$

$I_2=\frac{3}{8}A$

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Question

$R_1=R_3=R_4= 3\Omega$

$R_2=6\Omega$

In the circuit above, find the current through .

Answer

First, find the total resistance of the circuit.

and are in parallel, so we find their equivalent resistance by using the following formula:

$\frac{1}{R_{1}}+\frac{1}R_{2}=\frac{1}{R_{1+2}}$

$\frac{1}{3}+\frac{1}{6}=\frac{1}{R_{1+2}}$

$\frac{3}{6}=\frac{1}{R_{1+2}}}$

$R_{1+2}=2\Omega$

Next, add the series resistors together.

$R_{1+2}+R_3+R_4=R_{total}$

$R_{total}=8\Omega$

Use Ohm's law to find the current in the system.

$9V=I\cdot 8\Omega$

$I=\frac{9}{8}A$

In series, all resistors will have the same current.

Thus, the current through is the same as through the rest of the circuit.

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Question

What is the total resistance of the circuit?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

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Question

What is the current flowing through ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

We will then find the total current of the circuit. This will also be the current of because this resistor is not in parallel with any others.

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

$I=\frac{15V}{(67/13)\Omega}$

$I=\frac{195}{67} amps$

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Question

$R_1=R_4=R_5=2\Omega$

$R2=5\Omega$

$R_3=6 \Omega$

What is the current flowing through ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three

Using

Using algebraic subsitution we get:

$I_1+I_1 \frac{R_1}{R_2}+I_1 \frac{R_1}{R_3}=I_{total}$

Solving for

$\frac{I_{Total}}{1+\frac{R_1}{R_2}+\frac{R_1}{R_3}}=I_1$

$\frac{195/67}{1+\frac{2}{5}+\frac{2}{6}}=I_1$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the current flowing through ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three

Using

Using algebraic subsitution we get:

$I_2+I_2 \frac{R_2}{R_1}+I_2 \frac{R_2}{R_3}=I_{total}$

Solving for

$\frac{I_{total}}{1+\frac{R_2}{R_1}+\frac{R_2}{R_3}}=I_2$

$\frac{195/67}{1+\frac{5}{2}+\frac{5}{6}}=I_2$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the current flowing through ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three

Using

Using algebraic subsitution we get:

$I_3+I_3 \frac{R_3}{R_1}+I_3 \frac{R_3}{R_2}=I_{total}$

Solving for

$\frac{I_{Total}}{1+\frac{R_3}{R_1}+\frac{R_3}{R_2}}=I_3$

$\frac{I_{Total}}{1+\frac{6}{2}+\frac{6}{5}}=I_3$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the voltage drop across ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

We will then determine the total current of the circuit.

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

$I=\frac{15V}{(67/13)\Omega}$

$I=\frac{195}{67} amps$

Once again, using

Where is the resistance of the resistor in question, we get

$V=\frac{195}{67}\cdot 2\Omega$

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Question

$R_1=R_4=R_5=2\Omega$

$R_2=5\Omega$

$R_3=6\Omega$

What is the voltage drop across ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

First, we need to find the total current of the circuit, we simply use:

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

$I=\frac{15V}{\frac{67}{13}\Omega}$

$I=\frac{195}{67} amps$

Because , and are in parallel,

$I_1+I_2+I_3=I_{total}$

Also, the voltage drop must be the same across all three

Using

Using algebraic subsitution we get:

$I_2+I_2 \frac{R_2}{R_1}+I_2 \frac{R_2}{R_3}=I_{total}$

Solving for

$\frac{I_{total}}{1+\frac{R_2}{R_1}+\frac{R_2}{R_3}}=I_2$

$\frac{195/67}{1+\frac{5}{2}+\frac{5}{6}}=I_2$

Using

$V=.672A\cdot5\Omega$

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Question

The current through is , has a resistance of $1.5\Omega$ . Determine the voltage of the battery.

Answer

Due to the "loop rule", the voltage increase in the battery will need to be equal to the voltage drop in the resistor.

$V_1=V_{R1}$

Use Ohm's law:

Combine equations:

$V_1=I_{R1}R_{R1}$

Plug in values:

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Question

If the battery has a voltage of , and has a resistance of $2\Omega$ , determine the current through .

Answer

Due to the "loop rule", the voltage increase in the battery will need to be equal to the voltage drop in the resistor.

$V_1=V_{R2}$

Use Ohm's law:

Combine equations:

$V_1=I_{R2}R_{R2}$

Plug in values:

$6=2*I_{R2}$

$I_{R2}=3A$

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Question

A $9\textup{ V}$ battery is placed in series with five $9\ \Omega$ resistors. Find the total resistance.

Answer

In series, resistance adds conventionally.

$5*9\Omega=45\Omega$

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