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MCAT Physical : Electricity and Magnetism

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Example Questions

Example Question #1 : Resistors And Resistance

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What is the net resistance of the circuit?

Possible Answers:

$2.5\Omega$

$9\Omega$

$11\Omega$

$2\Omega$

$4\Omega$

Correct answer:

$4\Omega$

Explanation:

First we need to combine the resistors in parallel.

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

Now that we have simplified the two resistors in parallel, we need to determine how the R_eq of the two parallel resistors and the resistance due to R₁ are arranged. We can see that these resistors are now in series, thus we can directly add their resistances together to get the overall resistance of the circuit.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Remember as a general rule, the equivalent resistance of resistors in series is an arithmetic sum of their individual resistances.

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Example Question #21 : Circuits

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9 V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

All resistors in series share the same __________, while all resistors in parallel share the same __________.

Possible Answers:

current . . . voltage

current . . . potential energy

voltage . . . current

potential energy . . . voltage

Correct answer:

current . . . voltage

Explanation:

As a common rule to all circuits, all resistors in series share the same current (due to conservation of charge) and all resistors in parallel share the same voltage. This information is very helpful in determining the voltage drops across sets of resistors and the current that flows through them.

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Example Question #592 : Mcat Physical Sciences

What is the internal resistance of a 25V battery that has a potential of 10V when a current runs through it?

Possible Answers:

$15\Omega$

$5\Omega$

$30\Omega$

$3\Omega$

Correct answer:

$5\Omega$

Explanation:

The equation for a battery with internal resistance is $V = {emf} - IR_{internal}$

Using this equation, we can solve for the resistance.

$R = \frac{{emf}-V}{I} = \frac{25V-10V}{3A} = 5\Omega$

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Example Question #13 : Resistors And Resistance

A 25V battery is connected to a circuit which contains two $5\Omega$ resistors connected in parallel. What is the current throughout the circuit?

Possible Answers:

2.5A

10A

Correct answer:

10A

Explanation:

We know the total voltage in the circuit is 25V . We need to determine the total resistance within the circuit. Since we know there are two resistors connected in parallel we use the following formula:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Two resistors with resistances of $5\Omega$ each:

$\frac{1}{R_{eq}}=\frac{1}{5\Omega}+\frac{1}{5\Omega} = \frac{2}{5}\Omega$

$R_{eq}=\frac{5}{2}=2.5\Omega$

Now we use another formula to determine the current using voltage and resistance:

V=IR

Rearranged to:

$I=\frac{V}{R}= \frac{25V}{2.5\Omega }=10A$

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Example Question #61 : Electricity And Magnetism

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9 V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What direction do electrons flow through the circuit?

Possible Answers:

Counterclockwise

Clockwise

Cannot be determined

Correct answer:

Counterclockwise

Explanation:

Remember that convention dictates that current flows in the direction of positive charge (protons), thus, electrons flow in the opposite direction. Also remember that the larger length on the battery symbol on the circuit diagram indicates that current flows in this direction. In the diagram below, we can see that current flows clockwise, thus electrons flow counterclockwise.

Screen_shot_2013-10-09_at_11.11.43_pm

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Example Question #2 : Current

Batteries and AC current are often used to charge a capacitor. A common example of capacitor use is in computer hard drives, where capacitors are charged in a specific pattern to code information. A simplified circuit with capacitors can be seen below. The capacitance of C₁ is 0.5 μF and the capacitances of C₂ and C₃ are 1 μF each. A 10 V battery with an internal resistance of 1 Ω supplies the circuit.

Pretext Question_2

What direction is current flowing through the circuit?

Possible Answers:

Clockwise

Cannot be determined

Counterclockwise

Correct answer:

Counterclockwise

Explanation:

Remember that convention dictates that current flows in the direction of positive charge (protons). Also remember that the larger length on the battery symbol on the circuit diagram indicates that current flows in this direction. In the diagram below, we can see that current flows counterclockwise.

Current_direction

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Example Question #2 : Current

Screen_shot_2013-10-09_at_11.08.15_pm

How much current flows through resistor R₄?

Possible Answers:

2.25A

3.0A

1.5A

2.85A

0.75A

Correct answer:

1.5A

Explanation:

First, we need to determine the voltage drop across R₄. Given that R_A and R₄ are in parallel, we know that the voltage drop across each is the same.

The voltage drop across R₁ can be calculated using the I_circuit (all the current generated by the battery’s potential difference must pass through R₁ because it is in direct series with the battery), and the resistance of the resistor.

I can be calculated by V = IR because we have the voltage drop across the circuit (9 V) and can calculate the equivalent resistance:

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V=IR allows us to find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25 A

Plugging this value in, we can find the voltage drop across R₁.

V_R1 = IR = (2.25 A)(2Ω) = 4.5 V

Now we can determine the voltage drop across the parallel resistors by subtracting the voltage drop across R₁ from the battery.

V_A= 9V – 4.5V = 4.5V

this value will be the same for either R_A or R₄. We can now use V = IR to determine the amount of current that flows through R₄.

V = IR

I = V/R = 4.5V/3Ω = 1.5A

Additionally, due to Kirchhoff’s first law, we know that current in must equal current out in a junction. Thus, because R₄ has 1.5 A of current, R_A must have 0.75A. Thus makes logical sense because electrons take the path of least resistance, meaning the resistor with the lowest resistance will have the greater current.

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Example Question #3 : Current

Screen_shot_2013-10-09_at_11.08.15_pm

What is the current that flows through the circuit?

Possible Answers:

6.5A

2.25A

1.0A

3.25A

4.5A

Correct answer:

2.25A

Explanation:

Because we are being asked the current that flows through the circuit, we can use the formula V = IR because we have the voltage drop across the circuit (9V) and can calculate the equivalent resistance.

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V = IR, we can find the current.

V = IR

I = V/R = 9V/4Ω = 2.25A

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Example Question #4 : Current

If the peak voltage across a $\small 2.0k\Omega$ resistor in an AC circuit is $\small 12V$ , what is the rms current across it?

Possible Answers:

3.0 mA

4.2mA

11.9 mA

8.5 mA

6.1 mA

Correct answer:

4.2mA

Explanation:

First, find the rms voltage across the resistor with the equation:

$V_p=V_{rms}\sqrt{2}$

The pure voltage is $\small 12V$ , allowing us to solve for the rms voltage.

$V_{rms}=\frac{V_p}{\sqrt 2}= \frac{12 V}{\sqrt 2} = 8.485 V$

Now we can use Ohm's law to find the rms current.

$V_{rms}=I_{rms}R$

$I_{rms}=\frac{V_{rms}}{R}$

$I_{rms}=\frac{8.485V}{2000\Omega}$

$I_{rms}=0.00424A=4.2mA$

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Example Question #6 : Current

A 12V battery is connected in a circuit containing two resistors in parallel. The resistors have resistances of $2.0 k\Omega$ and $4.0 k\Omega$ . The $4.0 k\Omega$ resistor is then unscrewed, disconnecting that parallel branch. How does the new current, I_f, across the $2.0 k\Omega$ resistor compare to the original current across that resistor, I_o?

Possible Answers:

$\frac{1}{2}I_{o} < I_{f} < I_{o}$

$2I_{o} < I_{f}$

$I_{f} < \frac{1}{2}I_{o}$

$I_{o} = I_{f}$

$I_{o} < I_{f} < 2I_{o}$

Correct answer:

$I_{o} = I_{f}$

Explanation:

Regardless of whether the $4k\Omega$ resistor is present, we can simply focus on the current loop containing the battery and $2k\Omega$ resistor.

For resistors in parallel, each component/branch will have the same voltage drop, but the current may vary. Initially, the voltage through the $2k\Omega$ resistor will be:

$12V=I_o(2k\Omega)\rightarrow I_o=6A$

After the $4.0 k\Omega$ resistor is removed, the voltage will remain unchanged.

$12V=I_f(2k\Omega)\rightarrow I_f=6A$

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MCAT Physical : Electricity and Magnetism

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Resistors And Resistance

Example Question #21 : Circuits

Example Question #592 : Mcat Physical Sciences

Example Question #13 : Resistors And Resistance

Example Question #61 : Electricity And Magnetism

Example Question #2 : Current

Example Question #2 : Current

Example Question #3 : Current

Example Question #4 : Current

Example Question #6 : Current

All MCAT Physical Resources

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