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MCAT Physical : Electricity and Magnetism

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Example Question #4 : Current

Consider the following circuit:

Circuit_1__switch_creating_a_short_

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

Water has been spilt on the circuit, occasionally creating a short in the circuit. The short is represented by the switch, which is closed when the circuit is shorted. What is the difference in power loss through the circuit when the switch is closed compared to being open?

Possible Answers:

2.4W

1.6W

4.3W

8.9W

6.7W

Correct answer:

4.3W

Explanation:

We will need to calculate total equivalent resistance for each of the two scenarios to calculate power losses.

SCENARIO 1: Switch Closed

When the switch is closed, a resistance-free path is created. This effectively reduces the current flow through the parallel resistors to zero. Therefore, we have a much simpler circuit, consisting only of R1 and R5. Since they are in series, we can simply add them:

$R_{eq} = R1+R5 = 5\Omega + 2\Omega = 7\Omega$

The expression for power is:

P=IV

Substituting Ohm's law for current, we get:

$P = (\frac{V}{R})V=\frac{V^2}{R}$

Plugging in our values, we get:

$P = \frac{(12V)^2}{7\Omega} = 20.57 W$

SCENARIO 2: Switch Open

Now that we have resistors in parallel, it will take two steps to condense the circuit. For the parallel part, we get:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} =\frac{1}{R2}+\frac{1}{R3+R4}= \frac{1}{3}+\frac{1}{5} = \frac{8}{15}$

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

Now, since everything is in series in this equivalent circuit, we can simply add it all up:

$R_{total} = 5+\frac{15}{8}+2 = 8\frac{7}{8}\Omega$

Again using the formula for power, we get:

$P = \frac{V^2}{R} = \frac{(12V)^2}{\frac{71}{8}\Omega} = 16.23 W$

Now we can calculate the difference between the two scenarios:

$\Delta P = P_1 - P_2 = 20.57W-16.23 W = 4.3 W$

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Example Question #5 : Current

A 12V battery is connected in series with a $4000\Omega$ resistor and a 6mH inductor. What is the current in this circuit after a long period of connection?

Possible Answers:

2kA

48kA

3mA

288A

72mA

Correct answer:

3mA

Explanation:

Relevant equations:

$V = IR + L\frac{dI}{dt}$

When the circuit is first connected, the current increases from zero to its final value. During this time as the current changes, the inductor has a voltage across it. After a long period, the current has built up to its maximum value.

After a long period, $\frac{dI}{dt}=0$ .

$V = IR + L\frac{dI}{dt}\rightarrow V=IR$

Plugging in our values for voltage and resistance, we can solve for the final current.

$I=\frac{V}{R}=\frac{12V}{4000\Omega}=0.003A=3mA$

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Example Question #31 : Circuits

An circuit contains a $100\Omega$ resistor, 250nF capacitor, and 40mH inductor in series. If this circuit is connected to an AC generator, what angular frequency would maximize the current flow?

Possible Answers:

$25\frac{rad}{s}$

$5\frac{rad}{s}$

$1000\frac{rad}{s}$

$10000\frac{rad}{s}$

$200\frac{rad}{s}$

Correct answer:

$10000\frac{rad}{s}$

Explanation:

Relevant equations:

$\omega _o = \frac{1}{\sqrt{LC}}$

The current is maximized when the power supply frequency, $\omega$ , equals the resonance frequency, $\omega _o$ . Plugging in the given inductance and capacitance yields:

$\omega _o = \frac{1}{\sqrt{(40*10^{-3}H)(250*10^{-9}C)}}=\frac{1}{\sqrt{1*10^{-8}}}=\frac{1}{1*10^{-4}}= 10,000 \frac{rad}{s}$

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Example Question #7 : Current

An circuit consists of a $75\mu H$ inductor, 25nF capacitor, $400\Omega$ resistor, and a voltage source with maximum voltage of 200V in series. Approximately what is the rms current in this circuit at resonance?

Possible Answers:

20A

0.5A

0.125A

0.4A

Correct answer:

Explanation:

Relevant equations:

$I_{max}=\frac{V_{max}}{Z}$

$Z = \sqrt{R^2+(X_L-X_C)}$

X_L = X_C at resonance

$I_{rms}=\frac{I_{max}}{\sqrt2}$

Step 1: Find impedance, , at resonance:

$Z = \sqrt{R^2+0}=\sqrt{400}=20\Omega$

Step 2: Calculate $I_{max}$ , using $V_{max}=200V$ and $Z = 20\Omega$ :

$I_{max}=\frac{200V}{20\Omega}=10A$

Step 3: Use $I_{max}=10A$ to calculate $I_{rms}$ :

$I_{rms}=\frac{10A}{\sqrt2}\approx7A$

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Example Question #1 : Capacitors And Dielectrics

Capacitors with capacitances of 3 μF, 7 μF and 10 μF are wired in parallel. What is the capacitance of the circuit?

it cannot be determined without knowing the resistance of the circuit
it cannot be determined without knowing the time constant in the circuit
20 μF
approximately 1.75 μF
none of these is correct

Possible Answers:

Correct answer:

Explanation:

Response 3 is the correct choice. Electrons will space themselves as far apart as possible, because of charge repulsion; therefore, in a parallel arrangement, they will jump onto each capacitor and “load it up.” The parallel capacitance is calculated by simply adding the individual values. The value would be about 1.75 μF if the three were connected in series, where the formula is $\dpi{100} \small \frac{1}{C_{total}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$ (reciprocal of total equals sum of reciprocal of each). The time constant of a circuit is given by the formula τ = RC, and it relates to how fast a capacitor can charge to full capacitance.

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Example Question #2 : Capacitors And Dielectrics

A $2\mu F$ and $6\mu F$ capacitor are connected in series with a 12V battery. What is the maximum charge stored on the $2\mu F$ capacitor?

Possible Answers:

$18\mu C$

$24 \mu C$

$96 \mu C$

$13.5 \mu C$

$4.5 \mu C$

Correct answer:

$18\mu C$

Explanation:

First find the equivalent capacitance of the two capacitors by adding their inverses.

$\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{6}$ $\rightarrow C_{eq} = \frac{6}{4} = 1.5\mu F$

Then, we can find the charge stored on this equivalent capacitor.

$Q_{eq}=C_{eq}V = 1.5 * 12 = 18\mu C$

For capacitors in series the charges on each must be equal, and also equal to the charge on the equivalent capacitor. The answer is $18\mu C$ .

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Example Question #1 : Capacitors And Dielectrics

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

How is the charge stored on the capacitor?

Possible Answers:

Unevenly distributed on the capacitor surface

Unevenly distributed inside the capacitor

Evenly distributed on the capacitor surface

Evenly distributed inside the capacitor

Correct answer:

Evenly distributed on the capacitor surface

Explanation:

Charge is evenly distributed on the surface of the capacitor. If we think back to electric force, we know that positive charges repel other positive charges and negative charges repel other negative charges; thus, the charges are evenly distributed to minimize the force between them. We can see how this looks in diagrammatic form below.

Charge_dist

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Example Question #41 : Circuits

Pretext Question_2

What is the equivalent capacitance of C₂ and C₃?

Possible Answers:

0.5μF

0.33μF

3μF

2μF

Correct answer:

0.5μF

Explanation:

First, we need to determine how these capacitors are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals:

$\frac{1}{C_{eq}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C_eq = 0.5μF

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Example Question #5 : Capacitors And Dielectrics

Pretext Question_2

What is the equivalent capacitance of the circuit?

Possible Answers:

3μF

2μF

4μF

1μF

Correct answer:

1μF

Explanation:

First, we need to determine how capacitors C₂ and C₃ are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals.

$\frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C₂₃ = 0.5μF

Next, we need to determine how we can find the C_eq by simplifying C₂₃ and C₁. We can see that C_eq and C₁are in parallel, thus we can directly add the individual capacitances.

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

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Example Question #6 : Capacitors And Dielectrics

Pretext Question_2

How long does it take to fully charge the capacitors of the circuit?

Possible Answers:

1 * 10⁵s

1 * 10³s

1 * 10⁶s

1 * 10⁴s

Correct answer:

1 * 10⁶s

Explanation:

In order to determine the time, we need to know the total charge stored on the capacitors. Remember that Q = CV, where Q is the total charge, C is the equivalent capacitance, and V is the voltage. We must first find the equivalent capacitance.

C₂ and C₃ are capacitors in series, while C₁ is in parallel.

$\frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C₂₃ = 0.5μF

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

Now we can plug in the C_eq and battery voltage to find the charge.

Q = (1μF)(10V) = 10μC

Additionally, we need to know the current the battery can provide (the charge per unit time). Knowing both the total charge and current will allow us to calculate the time. We can use V = IR to determine the current.

I = V/R = 10V/1Ω = 10A = 10C/sec

We can equate charge and current to determine time.

10μC = 10 C/t

t = 10 C/10 μC = 1 * 10⁶s or 11.6days

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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 #4 : Current

Example Question #5 : Current

Example Question #31 : Circuits

Example Question #7 : Current

Example Question #1 : Capacitors And Dielectrics

Example Question #2 : Capacitors And Dielectrics

Example Question #1 : Capacitors And Dielectrics

Example Question #41 : Circuits

Example Question #5 : Capacitors And Dielectrics

Example Question #6 : Capacitors And Dielectrics

All MCAT Physical Resources

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