AP Physics 1 - Current and Voltage

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?

Explanation

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

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

Parallel circuit

What is the current in the given 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,

$V=IR_{eq}\Rightarrow$

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

Consider the circuit:

Circuit_4

How much current is flowing through R1?

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

Explanation

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:

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

Basic circuit

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?

Explanation

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

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

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

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

Explanation

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.

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

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

$R1=3k\Omega$

$R2=R3=4k\Omega$

$V_{SOURCE}=10V$

Circuit

What is the current passing though R3?

Explanation

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:

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

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

$R_{equivalent}=5k\Omega$

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.

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 .

Explanation

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 .

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

Consider the circuit:

Circuit_4

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

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

Explanation

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:

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

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

Wire B: and

Wire A: and

Wire C: and

Wire D: and

They would all have equal resistance

Explanation

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.

Consider the following circuit:

Circuit_1

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

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

Explanation

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:

Circuit_1__current_at_juntion_1_