Current and Voltage - AP Physics 1

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Question

Consider the following circuit:

What is the current flowing through R5?

$R1 = 5\Omega,\ R2 = 3\Omega,\ R3 = 4\Omega,\ R4 = 1\Omega,\ R5 = 2\Omega$

Answer

All of the current flowing through the circuit will go through R5, since it is not in parallel with anything. To find the current flowing through the circuit, we will need to first find the total equivalen resistance of the circuit.

To do this, we first we need to condense R3 and R4. They are in series, so we can simply add them to get:

$R_{34} = 5\Omega$

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{3}+\frac{1}{5} = \frac{8}{15}$

$R_{eq}=\frac{15}{8}$

The equivalent circuit now looks like:

Since everything is in parallel, we can simply add everything up:

$R_{total} = R1 + R_{eq}+ R5 = 5\Omega+\frac{15}{8}\Omega+2\Omega$

$R_{total} = 8\frac{7}{8}\Omega$

Now that we have the total resistance of the circuit, we can use Ohm's law to find the total current:

Rearranging for current, we get:

$I = \frac{V}{R} = \frac{12}{\frac{71}{8}} = \frac{96}{71}=1.35 A$

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Question

Consider the following circuit:

If 1.35A is flowing through R1, how much current is flowing through R4?

$R1=5\Omega,\ R2=3\Omega,\ R3=4\Omega,\ R4=1\Omega,\ R5=2\Omega$

Answer

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

Using Ohm's law to expand the voltages, we get:

$I_{2}R_{2} = I_{3}R_{3}+I_3R_4 = I_3(R_3+R_4)$

We now have two equations that we can solve simultaneously:

$I_{2}R_{2} =I_3(R_3+R_4)$

Since we are solve for I3, let's rearrange the first equation for I2:

Substituting this into the second equation, we get:

Rearranging to solve for I3:

$I_3 = \frac{I_1R_2}{R_2+R_3+R_4}$

We have all the values we need, so simply plug in the given values and solve:

$I_3 = \frac{(1.35)(3)}{(3+4+1)}= \frac{4.05}{8} = 0.51 A$

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Question

Consider the following circuit:

If 1.35 A is flowing through R1, what is the voltage drop across R2?

$R1=5\Omega,\ R2=3\Omega,\ R3=4\Omega,\ R4=1\Omega,\ R5=2\Omega$

Answer

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

Using Ohm's law to expand the voltages, we get:

$I_{2}R_{2} = I_{3}R_{3}+I_3R_4 = I_3(R_3+R_4)$

We now have two equations that we can solve simultaneously:

$I_{2}R_{2} =I_3(R_3+R_4)$

To find V2, we need to calculate I2, so let's rearrange the first equation for I3

Substituting this into the second equation, we get:

Rearranging for I2:

$I_2 = \frac{I_1(R_3+R_4)}{R_2+R_3+R_4}$

We have all the values we need, so simply plug in the given values and solve:

$I_2 = \frac{(1.35)(4+1)}{(3+4+1)}= \frac{6.75}{8} = 0.84 A$

Now that we know I2, we can use Ohm's law to find the voltage drop across that resistor:

$V_2 = I_2R_2 = (0.84A)(3\Omega) = 2.53 V$

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Question

Consider the following circuit:

By how much does the current flowing through the circuit increase when the switch is closed?

$R1 = 5\Omega,\ R2=3\Omega,\ R3=4\Omega,\ R4=1\Omega,\ R5=2\Omega$

Answer

In order to calculate the current flowing through the circuit in each scenario, we need to calculate the equivalent resistance in each scenario.

SCENARIO 1: Switch open

When the switch is open, everything is series, so we can simply add all of the resistances up to find the total equivalent resistance. Note that R2 will be excluded.

$R_{total} = R1+R3+R4+R5 = 5\Omega+4\Omega+1\Omega+2\Omega = 12\Omega$

Using Ohm's Law, we can now calculate the current flowing through the circuit:

Rearranging for current:

$I=\frac{V}{R}=\frac{12V}{12\Omega} = 1A$

SCENARIO 2: Switch closed

The circuit now looks like this:

Since we have a few resistors in parallel, this calcluation will involve a few more steps.

First we need to condense R3 and R4. They are in series, so we can simply add them to get:

$R_{34} = 5\Omega$

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{3}+\frac{1}{5} = \frac{8}{15}$

$R_{eq}=\frac{15}{8}$

The equivalent circuit now looks like:

Since everything is in series, we can simply add everything up:

$R_{total} = R1 + R_{eq}+ R5 = 5\Omega+\frac{15}{8}\Omega+2\Omega$

$R_{total} = 8\frac{7}{8}\Omega$

Now using Ohm's law to calculate the current flowing through the circuit:

$I = \frac{V}{R} = \frac{12}{\frac{71}{8}}=\frac{96}{71}=1.35A$

Now, we simply take the different of the currents in the two scenarios:

$\Delta I = 1.35 A - 1 A = 0.35 A$

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Question

Consider the circuit:

By how much does the current through the circuit decrease if R3 is removed?

$R1 = 5\Omega,\ R2 = 20\Omega,\ R3 = 25\Omega,\ R4 = 10\Omega$

Answer

We are asked to compare two different scenarios, each involving the calculation of equivalent resistance, which will use the following formula:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}$

Scenario 1: With R3 Present

$\frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{25\Omega}+\frac{1}{10\Omega}$

$\frac{1}{R_{eq}} = \frac{39}{100}\Omega$

$R_{eq} = 2.56\Omega$

Now using Ohm's law:

$I = \frac{V}{R}=\frac{12V}{2.56\Omega}=4.68A$

Scenario 2: Without R3

$\frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{10\Omega}$

$\frac{1}{R_{eq}} = \frac{7}{20}\Omega$

$R_{eq} = 2.86\Omega$

Now using Ohm's law:

$I = \frac{V}{R}=\frac{12V}{2.86\Omega}=4.20A$

Calculate the change in current:

$\Delta I = 4.68A-4.20A = 0.48A$

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Question

Consider the circuit:

How much current is flowing through R1?

$R1 = 2\Omega,\ R2 = 4\Omega,\ R3 = 4\Omega,\ R4 = 2\Omega$

Answer

Although it is possible to solve this problem by calculating an equivalent resistance, calculating a total current through the circuit, and then using Kirchoff's junction rule to find the current through R1, it is much easier to simply use Kirchoff's Loop rule. What this rule says is that through any closed loop in a circuit, all voltages must add up to zero. Written as an equation:

$\sum V = 0$

If we consider the closed loop path consisting of only the power supply and R1, we can use Ohm's law to calculate the current:

$I = \frac{12V}{2\Omega} = 6A$

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Question

Consider the circuit:

The total current through the circuit is and the current through R2 is . What is the value of R2?

Answer

The total current through the circuit is unnecessary to solve this problem. We only need to know the current through the resistor, as well as Kirchoff's loop rule. The loop rule states that for any closed loop through a circuit, the voltages add up to zero.

$\Delta V = 0$

For this problem, we will consider the loop that consists solely of the voltage source and R2. From the rule, we know that 12V is lost through the resistor. Using Ohm's law, we can write:

Rearranging for R2, we get:

$R2 = \frac{12V}{I_2} = \frac{12V}{7A}= 1.7\Omega$

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Question

If a TV uses of energy over the course of , and it has a voltage of , how many coulombs passed through it during that time?

Answer

Because the TV uses , and it was used for , it must have used

.

$Power = Voltage \cdot Current$ so:

, and since the TV was used for

$3 :coulombs/sec \cdot 900: sec = 2700 :coulombs$

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Question

A student has created the given circuit diagram. It consists of a battery, a resistor, and a light bulb. In one minute, 1.2C of charge flows through the resistor. How much charge flows through the light bulb in one minute?

Answer

Since the bulb and the resistor are connected in series, the current is the same in each. Electric current is just the flow of charge through the circuit, so the same amount of charge flows through each in one minute.

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Question

In this circuit above, $Z_1=4\Omega$ and $Z_2=12\Omega$ . The voltage drop across is eight volts. What is the current across the circuit?

Answer

The voltage drop is related to the current and resistance via Ohm's law:

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

$I=\frac{8V}{4\Omega}$

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Question

Give the SI units for voltage.

Answer

We have to know what SI units are for this problem. SI units are units that cannot be broken down into smaller component parts. Voltage is defined as:

$V=\frac{J}{C}$ , where is joules and is Coulombs. Coulombs are already in SI units. However, joules is composed of other terms.

$J=N\cdot m=\left(\frac{kg\cdot m}{s^2}\right)\cdot m=\frac{kg\cdot m^2}{s^2}$

Therefore, voltage can be written as

$V=\frac{kg\cdot m^2}{s^2\cdot C}$

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Question

$5.6\times 10^{8}C$ of charge pass through a television during a movie. How much is being used by the television?

Answer

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

$2hr\times \frac{60min}{1hr}\times \frac{60sec}{1min}=7200 sec$

Solving for current,

$I=\frac{Q}{t}=\frac{5.6\times 10^{8}}{7200}=7.78\times10^{4}A$

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Question

Who long (in hours) does it take a computer to use $7.8\times10^{6}C$ of charge $\left (I=5A \right )$ ?

Answer

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

For this problem

$t=\frac{Q}{I}=\frac{7.8\times10^{6}}{5}=1.5\times10^{6}sec\times \frac{1min}{60sec}\times \frac{1hr}{60min}=433hr$

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Question

What quantity of charge passes through a clock in if its current is ?

Answer

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

$30min\times \frac{60sec}{1min}=1800 sec$

Solving for charge,

$Q=It=5\cdot1800=9000C$ .

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Question

You have the following circuit. The values for the components are:

$R1=3k\Omega$

$R2=R3=4k\Omega$

$V_{SOURCE}=10V$

What is the current passing though R3?

Answer

The first step is to find the equivalent resistance of the entire circuit. Note that R2 and R3 are in parallel, and they are in series with R1. Therefore, in order to find the equivalent resistance:

$(R2\left | \right |R3)+R1=\frac{R2R3}{R2+R3}+R1=\frac{4k\Omega4k\Omega}{4k\Omega+4k\Omega}+3k\Omega$

$R_{equivalent}=2k\Omega+3k\Omega$

$R_{equivalent}=5k\Omega$

Then, in order to find the total current, use Ohm's Law:

$V=IR\rightarrow I=\frac{V}{R}\rightarrow I=\frac{10}{5}A=2A$

That means that will pass through R1, and will split when it reaches the junction between R2 and R3 based on the ratio of the two resistance:

$I_{R3}=\frac{R2}{R2+R3}I_{source}$

$I_{R3}=\frac{1}{2}(2A)=1A$

of current passes through the resistor R3.

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Question

A circuit with one battery and one resistor with resistance $1 \Omega$ has current flowing through it. It is modified to add two more identical resistors in series of the first resistor and 1 more resistor in parallel to the other three with resistance $3\Omega$ . Determine the current flowing through the original resistor.

Answer

To find our answer, we have to know how each of the new additions changes the circuit. The addition of the two resistors in series do not affect the current because current is equivalent for resistors in series. However, since there is a resistor in parallel with equivalent resistance to the sum of the 3 resistors in series, current will split evenly amongst the two pathways, thus leading to a current of $\frac{I_0}{2}$ flowing through the original resistor.

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Question

Consider a circuit composed of a battery and three resistors in series. The three resistors are $R_1 = 3 \Omega, R_2 = 5 \Omega, R_3 = 8 \Omega$ . Calculate the voltage drop $(\Delta V)$ across .

Answer

In order to find the voltage drop $(\Delta V)$ across , we will use Ohm's law:

$\Delta V=IR$ .

However, we first need to solve for the current in the circuit. This will be calculated by applying Ohm's law to the entire circuit, using the total resistance $(R_{tot})$ for resistors in series.

$V=IR\Rightarrow I = \frac{V}{R_{tot}}=\frac{18V}{16 \Omega}=1.125 A$

Finally, we use this calculated current to find the voltage drop across .

$\Delta V_2=IR_2=(1.125A)5\Omega=5.625V$

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Question

Determine the voltage drop across a resistor of $2\Omega$ experiencing a current flowing through it, if it is connected to a battery of ?

Answer

In circuits with resistors, the only thing necessary to determine voltage drop across a resistor is the current through it and the resistance, as given by Ohm's law.

$\Delta V=IR=2A*2\Omega=4V$

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Question

Which of the following wires with a current running through them would have the least resistance?

Answer

The correct answer is Wire B with and of current.

The formula for voltage, current and resistance is as follows:

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

Thus, the wire with the lowest ratio of voltage to current will have the least resistance.

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Question

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

Answer

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

$R_{eq}=R_1+R_2+...$

For this problem

$R_{eq}=R_1+R_2=10+15=25\Omega$

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

$V=IR_{eq}\Rightarrow I=\frac{V}{R_{eq}}=\frac{25}{25}=1A$

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