Practice Questions - Circuits - MCAT Physics and Math Review

MCAT Physics and Math Review

Chapter 6: Circuits

Practice Questions

1. If a defibrillator passes 15 A of current through a patient’s body for 0.1 seconds, how much charge goes through the patient’s skin?

1. 0.15 C

2. 1.5 C

3. 15 C

4. 150 C

2. A charge of 2 μC flows from the positive terminal of a 6 V battery, through a 100 Ω resistor, and back through the battery to the positive terminal. What is the total potential difference experienced by the charge?

1. 0 V

2. 0.002 V

3. 0.2 V

4. 6 V

3. The resistance of two conductors of equal cross-sectional area and equal lengths are compared, and are found to be in the ratio 1:2. The resistivities of the materials from which they are constructed must therefore be in what ratio?

1. 1:1

2. 1:2

3. 2:1

4. 4:1

4. A voltaic cell provides a current of 0.5 A when in a circuit with a 3 Ω resistor. If the internal resistance of the cell is 0.1 Ω, what is the voltage across the terminals of the battery when there is no current flowing?

1. 0.05 V

2. 1.5 V

3. 1.505 V

4. 1.55 V

5. A transformer is a device that takes an input voltage and produces an output voltage that can be either larger or smaller than the input voltage, depending on the transformer design. Although the voltage is changed by the transformer, energy is not, so the input power equals the output power. A particular transformer produces an output voltage that is 300 percent of the input voltage. What is the ratio of the output current to the input current?

1. 1:3

2. 3:1

3. 1:300

4. 300:1

6. Given that R1 = 20 Ω, R2 = 4 Ω, R3 = R4 = 32 Ω, R5 = 15 Ω, and R6 = 5 Ω, what is the total resistance in the circuit shown below?

1. 0.15 Ω

2. 6.67 Ω

3. 16.7 Ω

4. 60 Ω

7. How many moles of electrons pass through a circuit containing a 100 V battery and a 2 Ω resistor over a period of 10 seconds? (Note: )

1. 5.18 × 10–3 moles

2. 500 moles

3. 5.18 × 103 moles

4. 5.2 × 106 moles

8. In the circuit below, what is the voltage drop across the resistor?



3. 5 V

4. 7.5 V

9. If the area of a capacitor’s plates is doubled while the distance between them is halved, how will the final capacitance (Cf) compare to the original capacitance (Ci)?

1. Cf = Ci


3. Cf = 2Ci

4. Cf = 4Ci

10.The energy stored in a fully charged capacitor is given by . In a typical cardiac defibrillator, a capacitor charged to 7500 V has a stored energy of 400 J. Based on this information, what is the charge on the capacitor in the cardiac defibrillator?

1. 1.1 × 10−5 C

2. 5 × 10−2 C

3. 1.1 × 10−1 C

4. 3.1 × 106 C

11.A 10 Ω resistor carries a current that varies as a function of time as shown. How much energy has been dissipated by the resistor after 5 s?

1. 40 J

2. 50 J

3. 80 J

4. 120 J

12.In the figure below, six currents meet at point P. What is the magnitude and direction of the current between points P and x?

1. 2 A, toward x

2. 2 A, toward P

3. 10 A, toward x

4. 10 A, toward P

13.Which of the following will most likely increase the electric field between the plates of a parallel plate capacitor?

1. Adding a resistor that is connected to the capacitor in series

2. Adding a resistor that is connected to the capacitor in parallel

3. Increasing the distance between the plates

4. Adding an extra battery to the system

14.Each of the resistors shown carries an individual resistance of 4 Ω. Assuming negligible resistance in the wire, what is the overall resistance of the circuit?

1. 16 Ω

2. 8 Ω

3. 4 Ω

4. 3 Ω

15.Which of the following best characterizes ideal voltmeters and ammeters?

1. Ideal voltmeters and ammeters have infinite resistance.

2. Ideal voltmeters and ammeters have no resistance.

3. Ideal voltmeters have infinite resistance, and ideal ammeters have no resistance.

4. Ideal voltmeters have no resistance, and ideal ammeters have infinite resistance.


Answers and Explanations

1. BElectrical current is defined as charge flow, or in mathematical terms, charge transferred per time: A 15 A current that acts for 0.1 s will transfer 15 A × 0.1 s = 1.5 C of charge.

2. AKirchhoff’s loop rule states that the total potential difference around any closed loop of a circuit is 0 V. Another way of saying this is that the voltage gained in the battery (6 V) will be used up through the resistors. Because this charge both started and ended at the positive terminal, its total potential difference is therefore 0 V. 6 V, choice (D), is the voltage gained in the battery as well as the voltage drop in the resistors—creating a net sum of 0 V.

3. BThe resistance of a resistor is given by the formula Thus, there is a direct proportionality between resistance and resistivity. Because the other variables are equal between the two resistors, we can determine that if R1:R2 is a 1:2 ratio, then ρ1: ρ2 is also a 1:2 ratio.

4. D

This question tests our understanding of batteries in a circuit. The voltage across the terminals of the battery when there is no current flowing is referred to as the electromotive force (emf or ε of the battery). However, when a current is flowing through the circuit, the voltage across the terminals of the battery is decreased by an amount equal to the current multiplied by the internal resistance of the battery. Mathematically, this is given by the equation

V = ε – irint

To determine the emf of the battery, first calculate the voltage across the battery when the current is flowing. For this, we can use Ohm’s law:

Because we know the internal resistance of the battery, the current, and the voltage, we can calculate the emf:

The answer makes sense in the context of a real battery because its internal resistance is supposed to be very small so that the voltage provided to the circuit is as close as possible to the emf of the cell when there is no current running.

5. AWe are told that transformers conserve energy so that the output power equals the input power. Thus Pout = Pin, or IoutVout = IinVin. There is therefore an inverse proportionality between current and voltage. If the output voltage is 300% of the input voltage (3 times its amount), then the output current must be of the input voltage. This can be represented as a 1:3 ratio.

6. B

The fastest way to tackle these kinds of questions is to simplify the circuit bit by bit. For example, notice that R3 and R4 are in parallel with each other and are in series with R2; similarly, R5 and R6 are in series. If we determine the total resistance in each branch, we will be left with three branches in parallel. To start, find the total resistance in the middle branch:

Next, take a look at the total resistance in the bottom branch:

R5+6 = R5 + R6 = 15 Ω + 5 Ω = 20 Ω

The circuit can now be viewed as three resistors in parallel, each providing a resistance of 20 Ω. The total resistance in the circuit is thus

7. A

To determine the moles of charge that pass through the circuit over a period of 10 s, we will have to calculate the amount of charge running through the circuit. Charge is simply current times time, and the current can be calculated using Ohm’s law:

Then, calculate the number of moles of charge that this represents by using the Faraday constant and approximating F as

This is closest to choice (A).

8. C

To determine the voltage drop across the resistor, start by calculating the total resistance in the circuit. For the resistors in parallel, the total resistance is

The total resistance in the circuit is the sum of the remaining resistor and the equivalent resistance of the other two:

Now that we know the equivalent resistance, we can calculate the total current using Ohm’s law:

Finally, we can determine the voltage drop across the parallel resistors. The voltage drop across the resistor must be Therefore, there must be a 5 V drop across both the resistor and 2 Ω resistor, according to Kirchhoff’s loop rule. Each of these resistors forms a complete loop in combination with the resistor and 10 V voltage source, and the net potential difference around any closed loop must be 0 V.

9. DThis question should bring to mind the equation where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. From this equation, we can infer that doubling the area will double the capacitance, and halving the distance will also double the capacitance. Therefore, the new capacitance is four times larger than the initial capacitance.


Because the question is asking us to calculate the charge on the capacitor, use the formula Q = CV. We are given V = 7500 V and can calculate C from the formula for energy, :

Thus, the charge is close to 0.1 C, which is closest to choice (C).


Power is energy dissipated per unit time; therefore, the energy dissipated is E = PΔt. In the five-second interval during which the resistor is active, it has a 2 A current for three of those seconds. The power dissipated by a resistor R carrying a current I is P = I2R. Therefore, the energy dissipated is

E = I2RΔt =(2 A)2(10 Ω)(3 s)= 4 × 10 × 3 = 120 J

12.AKirchhoff’s junction rule states that the sum of all currents directed into a point is always equal to the sum of all currents directed out of the point. The currents directed into point P are 8 A, 2 A, and 3 A, so the sum is 13 A. The currents directed out of point P are 5 A and 6 A, so the total is 11 A. Because the two numbers must always be equal, an additional current of 2 A must be directed away from point P toward point x.

13.DThe electric field between two plates of a parallel plate capacitor is related to the potential difference between the plates of the capacitor and the distance between the plates, as shown in the formula The addition of another battery will increase the total voltage applied to the circuit, which, consequently, is likely to increase the electric field. The addition of a resistor in series will increase the resistance and decrease the voltage applied to the capacitor, eliminating choice (A). Adding a resistor in parallel will not change the voltage drop across the capacitor and should not change the electric field, eliminating choice (B). Increasing the distance between the plates, choice (C), would decrease the electric field, not increase it.


The resistance of the three resistors wired in series is equal to the sum of the individual resistances (12 Ω). This means that the circuit essentially contains a 12 Ω resistor and a 4 Ω resistor. To determine the overall resistance of this system, use the formula

15.CWhile this is primarily a recall question, it should also be intuitive. Voltmeters are attempting to determine a change in potential from one point to another. To do this, they should not provide an alternate route for charge flow and should therefore have infinite resistance. Ammeters attempt to determine the flow of charge at a single point and should not contribute to the resistance of a series circuit; therefore, they should have no resistance.