Current and Voltage

Help Questions

AP Physics 2 › Current and Voltage

Questions 1 - 10

1

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Determine the voltage drop across

None of these

Explanation

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.

.

2

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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 :

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

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

3

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.

None of these

Explanation

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

4

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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:

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

5

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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:

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

6

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Explanation

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:

$V=1.14A*2\Omega$

7

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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:

8

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Explanation

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:

$V=2.57A*3\Omega$

9

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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:

10

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the current through .

Explanation

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:

.

Page 1 of 2

Return to subject