Current and Voltage - AP Physics 2

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Question

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

Answer

The relevant equation for this problem is Ohm's law.

We're given the voltage and the current, so we need to rearrange to get the resistance.

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

$R=\frac{9}{1.5}$

$R= 6\Omega$

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Question

Consider a circuit connected to a battery that has a resistance of $5 \Omega$ . How many moles of charge pass through this circuit within a 30s period?

$F=9.65*10^{4}\frac{C}{mol}$

Answer

To begin with, we'll need to calculate the current that is flowing through this circuit. We can do this by using Ohm's law.

$I=\frac{V}{R}=\frac{180V}{5 \Omega }=36\frac{C}{s}$

Now that we have an expression that gives us the current, which is in units of charge per second, we'll need to calculate the moles of charge per second.

$I=\frac{Q}{t}$

$Q=It=36 \frac{C}{s}(30s)=1080C$

This expression directly above gives the total amount of charge that passes within a given time frame. But to find the moles of charge, we'll need to use Faraday's constant.

$moles\ of\ charge=\frac{Q}{F}$

$moles:of:charge=1080C\left ( \frac{1mol}{9.65* 10^{4} C} \right )$

$moles:of:charge=1.12*10^{-2}: mol$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohms law to determine the total current of the circuit

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega }=3.43 A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

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

Solve for :

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

$\frac{3.43A}{(1+\frac{1\Omega }{2\Omega })}=2.29A$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit:

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

Combine all of our voltage sources:

Plug in our values:

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values:

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohms law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plug in our values:

$\frac{3.43 A}{(1+\frac{3 \Omega}{1\Omega})}=I_3$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plugin our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_4(1+\frac{R_4}{R_3})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_4}{R_3})}=I_4$

$\frac{3.43amps}{(1+\frac{1\Omega}{3\Omega})}=I_4$

Plug in our values:

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solve for :

$\frac{3.43A}{(1+\frac{2\Omega}{3\Omega})}=I_5$

Plug in our values:

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

$\frac{3.43A}{(1+\frac{3\Omega}{2\Omega})}=I_6$

Plug in our values:

.

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit

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

Combine all of our voltage sources:

Plug in our values:

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values.

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

Use Ohm's law:

$V=1.14A*2\Omega$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plug in our values:

$\frac{3.43 A}{(1+\frac{3 \Omega}{1\Omega})}=I_3$

Use Ohm's law:

$V=.856A*3\Omega$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_4(1+\frac{R_4}{R_3})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_4}{R_3})}=I_4$

$\frac{3.43amps}{(1+\frac{1\Omega}{3\Omega})}=I_4$

Plug in our values, we get

Use:

$V=2.57A*3\Omega$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solve for :

$\frac{3.43A}{(1+\frac{2\Omega}{3\Omega})}=I_5$

Plug in our values, we get

Use:

$V=2.06A*2\Omega$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

$\frac{3.43A}{(1+\frac{3\Omega}{2\Omega})}=I_6$

Plug in our values, we get

Use

$V=1.37a*3\Omega$

Compare your answer with the correct one above

Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Determine the voltage drop across

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining with , with , with .

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

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

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

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

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

Solving for :

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

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

Plugging back into ohms law in order to find the voltage drop.

.

Compare your answer with the correct one above

Question

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the current through each resistor.

Answer

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

It is determined that

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Question

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the resistance of each resistor.

Answer

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

It is determined that

Using

$R=3\Omega$ for all three resistors

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Question

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

Answer

The relevant equation for this problem is Ohm's law.

We're given the voltage and the current, so we need to rearrange to get the resistance.

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

$R=\frac{9}{1.5}$

$R= 6\Omega$

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Question

Consider a circuit connected to a battery that has a resistance of $5 \Omega$ . How many moles of charge pass through this circuit within a 30s period?

$F=9.65*10^{4}\frac{C}{mol}$

Answer

To begin with, we'll need to calculate the current that is flowing through this circuit. We can do this by using Ohm's law.

$I=\frac{V}{R}=\frac{180V}{5 \Omega }=36\frac{C}{s}$

Now that we have an expression that gives us the current, which is in units of charge per second, we'll need to calculate the moles of charge per second.

$I=\frac{Q}{t}$

$Q=It=36 \frac{C}{s}(30s)=1080C$

This expression directly above gives the total amount of charge that passes within a given time frame. But to find the moles of charge, we'll need to use Faraday's constant.

$moles\ of\ charge=\frac{Q}{F}$

$moles:of:charge=1080C\left ( \frac{1mol}{9.65* 10^{4} C} \right )$

$moles:of:charge=1.12*10^{-2}: mol$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohms law to determine the total current of the circuit

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega }=3.43 A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

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

Solve for :

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

$\frac{3.43A}{(1+\frac{1\Omega }{2\Omega })}=2.29A$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

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

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

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit:

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

Combine all of our voltage sources:

Plug in our values:

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values:

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

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