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MCAT Physical : Resistors and Resistance

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Example Questions

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

voltage . . . current

current . . . potential energy

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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MCAT Physical : Resistors and Resistance

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #21 : Circuits

Example Question #592 : Mcat Physical Sciences

Example Question #13 : Resistors And Resistance

All MCAT Physical Resources

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