MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4c Resistors Capacitors Series Parallel

What this quiz covers

This quiz focuses on 4c Resistors Capacitors Series Parallel, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It becomes larger than $20\ \text{nF}$ because series capacitors add directly.
2. It becomes $5\ \text{nF}$ because equal capacitors in series halve the capacitance. (correct answer)
3. It becomes $20\ \text{nF}$ because the total is conserved when components are split.
4. It becomes $10\ \text{nF}$ because equal capacitors in series double the capacitance.

Explanation: This question tests understanding of replacing a single capacitor with series capacitors. For capacitors in series, 1/C_eq = 1/C₁ + 1/C₂. With two 10 nF capacitors in series, 1/C_eq = 1/10 + 1/10 = 2/10, giving C_eq = 5 nF. This is one-quarter of the original 20 nF capacitor, not half, because each replacement capacitor is already half the original value. The correct answer recognizes that equal capacitors in series result in half the capacitance of a single capacitor (10 nF → 5 nF). Choice A incorrectly applies series addition. Choice C incorrectly assumes conservation of total capacitance. The key insight is that series capacitors always reduce total capacitance, and using smaller individual capacitors further reduces the result.

Question 2

1. Charging becomes faster because the series resistance decreases. (correct answer)
2. Charging becomes slower because the capacitor now has less voltage across it.
3. Charging is unchanged because resistors only affect steady-state current, not transients.
4. Charging becomes faster because the equivalent capacitance increases when a resistor is removed.

Explanation: This question tests understanding of how series resistance affects RC charging rates. Initially, the total resistance is R₁ + R₂ = 10k + 10k = 20 kΩ, giving τ_initial = 20k × 1μ = 20 ms. After shorting R₂, only R₁ = 10 kΩ remains, giving τ_final = 10k × 1μ = 10 ms. A smaller time constant means faster charging, as the capacitor reaches 63.2% of final voltage in less time. The correct answer recognizes that reducing series resistance decreases the RC time constant and speeds up charging. Choice B incorrectly focuses on voltage division rather than charging rate. Choice C incorrectly claims resistors don't affect transients. The practical insight is that reducing series resistance in an RC circuit always speeds up the transient response.

Question 3

1. The capacitor initially behaves like an open circuit, so the divider output jumps to its final value instantly.
2. The capacitor initially behaves like a short circuit, pulling the divider output toward the lower node before settling. (correct answer)
3. The capacitor increases the series resistance, slowing the current through both resistors equally.
4. The capacitor decreases the total series resistance permanently, changing the final divider ratio.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. When voltage is first applied to an RC circuit, an uncharged capacitor initially acts like a short circuit, allowing maximum current flow. In this voltage divider, the capacitor parallel to R2 initially shorts out R2, pulling the divider output toward ground (0V) before the capacitor charges and the output settles to its steady-state value determined by the resistor ratio. The correct answer recognizes this initial short-circuit behavior of capacitors. Answer A incorrectly describes initial behavior as open circuit, while answers C and D make incorrect claims about the circuit's behavior.

Question 4

1. Total capacitance increases, allowing more charge to be stored at the same voltage. (correct answer)
2. Total capacitance decreases, allowing less charge to be stored at the same voltage.
3. Total resistance increases, so less charge is stored at the same voltage.
4. Total resistance decreases, so the voltage across each capacitor must decrease.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Capacitors in parallel add directly: C_total = C1 + C2, so two 1 μF capacitors in parallel give 2 μF total capacitance. Since Q = CV, doubling the capacitance at the same voltage doubles the charge storage capability. The correct answer recognizes that parallel capacitors increase total capacitance and charge storage. Answer B incorrectly states capacitance decreases, while answers C and D incorrectly invoke resistance concepts when the question focuses on capacitor behavior.

Question 5

1. Series capacitors yield a larger effective capacitance than either capacitor alone.
2. Series capacitors yield a smaller effective capacitance than either capacitor alone. (correct answer)
3. Series capacitors behave like series resistors, so effective capacitance is the sum.
4. Effective capacitance is unchanged because capacitance depends only on voltage.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Capacitors in series combine according to 1/C_total = 1/C1 + 1/C2, so two 5 pF capacitors in series yield C_total = 2.5 pF, which is smaller than either individual capacitor. This reduction in capacitance is desirable for modeling myelinated axons where reduced capacitance speeds up voltage changes (faster time constant). The correct answer identifies that series capacitors yield smaller effective capacitance. Answer A incorrectly states capacitance increases, answer C incorrectly applies resistor addition rules to capacitors, and answer D incorrectly claims capacitance is unchanged.

Question 6

1. Branch 2 current is nonzero because the series resistor allows DC through the capacitor.
2. Branch 2 current goes to zero because the capacitor blocks DC at steady state. (correct answer)
3. Branch 2 current increases over time because the capacitor gradually becomes a better conductor.
4. Branch 2 current equals Branch 1 current because the branches are in parallel.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. At DC steady state, capacitors act as open circuits and block all DC current. In Branch 2, the series capacitor prevents any steady-state current flow regardless of the resistor value, so the branch current goes to zero. The correct answer recognizes that capacitors block DC at steady state. Answer A incorrectly claims the resistor allows DC through the capacitor, answer C incorrectly states current increases over time, and answer D incorrectly claims equal currents in both branches.

Question 7

1. Voltage decays faster because the effective resistance is lower, increasing leakage. (correct answer)
2. Voltage decays slower because lower resistance reduces discharge current.
3. Voltage decays faster because the effective capacitance decreases in parallel.
4. Voltage is unaffected because capacitors determine only steady-state behavior.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. In the parallel RC membrane model, decreasing Rm (opening more leak channels) provides a lower resistance discharge path for the capacitor. With τ = RmCm, a smaller Rm means a smaller time constant and faster voltage decay after a current pulse. The correct answer recognizes that lower resistance increases leakage and speeds voltage decay. Answer B incorrectly states decay is slower, answer C incorrectly attributes the effect to capacitance changes, and answer D incorrectly claims voltage is unaffected.

Question 8

1. It is 9 V because the capacitors in parallel share the same voltage as the battery they were charged with. (correct answer)
2. It is 18 V because the two capacitor voltages add in parallel.
3. It is 4.5 V because the voltage splits between the two capacitors in parallel.
4. It is 0 V because capacitors cannot provide current to a resistor.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Capacitors in parallel share the same voltage, so both capacitors charge to the battery voltage of 9V. When disconnected from the battery and connected to a resistor, the capacitor pair initially maintains this 9V across the resistor before beginning to discharge. The correct answer recognizes that parallel capacitors share the same voltage. Answer B incorrectly adds voltages, answer C incorrectly divides voltage, and answer D incorrectly claims capacitors can't provide current.

Question 9

1. The equivalent resistance is $4\ \text{M}\Omega$ because parallel resistances add directly.
2. The equivalent resistance is $1\ \text{M}\Omega$, which increases the current drawn from the electrode compared with a single $2\ \text{M}\Omega$ resistor. (correct answer)
3. The equivalent resistance is $2\ \text{M}\Omega$ because identical resistors in parallel behave like a single resistor of the same value.
4. The equivalent resistance is $0.5\ \text{M}\Omega$ because resistances in parallel halve twice for two components.

Explanation: This question tests understanding of resistors in parallel configurations. When resistors are connected in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of individual resistances: 1/Req = 1/R1 + 1/R2. For two identical 2 MΩ resistors in parallel, 1/Req = 1/2 + 1/2 = 1 MΩ^(-1), giving Req = 1 MΩ. This reduced resistance increases the current drawn from the electrode compared to a single 2 MΩ resistor, as current = voltage/resistance. Choice A incorrectly adds resistances directly, which only applies to series circuits. A useful check is remembering that parallel resistance is always less than the smallest individual resistance.

Question 10

1. The equivalent capacitance is smaller than $1\ \mu\text{F}$, so less total charge is stored for the same applied voltage. (correct answer)
2. The equivalent capacitance is larger than $1\ \mu\text{F}$, so more total charge is stored for the same applied voltage.
3. The equivalent capacitance equals $2\ \mu\text{F}$ because capacitances in series add directly.
4. The equivalent capacitance equals $1\ \mu\text{F}$ because series connection does not change capacitance for identical components.

Explanation: This question tests understanding of capacitors in series configurations. When capacitors are connected in series, the reciprocal of the equivalent capacitance equals the sum of the reciprocals of individual capacitances: 1/Ceq = 1/C1 + 1/C2. For two identical 1 μF capacitors in series, 1/Ceq = 1/1 + 1/1 = 2 μF^(-1), giving Ceq = 0.5 μF. This reduced capacitance means less charge is stored at the same voltage, since Q = CV. Choice C incorrectly adds capacitances directly, which only applies to parallel circuits. Remember that series capacitance is always less than the smallest individual capacitance, opposite to the behavior of resistors.

Question 11

1. The equivalent capacitance is greater than $1\ \mu\text{F}$ because series capacitors add.
2. The equivalent capacitance is less than $1\ \mu\text{F}$ because series combination is smaller than the smallest capacitor. (correct answer)
3. The equivalent capacitance equals $3\ \mu\text{F}$ because $1+2=3$ in series.
4. The equivalent capacitance equals $2\ \mu\text{F}$ because the larger capacitor dominates in series.

Explanation: This question tests understanding of capacitors in series configurations. For capacitors in series, the reciprocal rule applies: 1/Ceq = 1/C1 + 1/C2. With C1 = 1 μF and C2 = 2 μF, we get 1/Ceq = 1/1 + 1/2 = 3/2 μF^(-1), giving Ceq = 2/3 μF ≈ 0.67 μF. This is less than the smallest capacitor (1 μF), which is always true for series capacitors. The series combination reduces the ability to store charge compared to either capacitor alone. Choice A incorrectly states series capacitors add, which only applies to parallel configurations. Remember that series capacitance is always smaller than the smallest individual capacitance.

Question 12

1. Total current decreases because the equivalent resistance increases when switching to parallel.
2. Total current increases because the equivalent resistance decreases when switching to parallel. (correct answer)
3. Total current is unchanged because the same resistors are used in both circuits.
4. Total current increases because equivalent capacitance increases when resistors are placed in parallel.

Explanation: This question tests understanding of how series versus parallel configurations affect total current draw. Circuit X has two 5 kΩ resistors in series: Req = 5 + 5 = 10 kΩ. Circuit Y has the same resistors in parallel: 1/Req = 1/5 + 1/5 = 2/5 kΩ^(-1), giving Req = 2.5 kΩ. Since current I = V/R at fixed voltage, Circuit Y with lower resistance (2.5 kΩ vs 10 kΩ) draws more current. The parallel configuration provides multiple current paths, reducing overall resistance and increasing current. Choice A incorrectly states resistance increases in parallel, while choice D incorrectly mentions capacitance in a purely resistive circuit. Remember: parallel always decreases resistance and increases current.

Question 13

1. Total resistance decreases, so the initial discharge current increases. (correct answer)
2. Total resistance increases, so the initial discharge current increases.
3. Total capacitance decreases, so the initial discharge current decreases.
4. Total capacitance increases, so the initial discharge current becomes zero.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Originally, the total resistance was R1 + R2 = 20 Ω + 30 Ω = 50 Ω, but replacing R2 with a wire (0 Ω) reduces total resistance to just R1 = 20 Ω. For a capacitor discharge, the initial current is I0 = V0/R, where V0 is the initial capacitor voltage. With decreased resistance (50 Ω → 20 Ω), the initial discharge current increases by a factor of 50/20 = 2.5. This creates a more rapid, higher-current discharge pulse. Choice B incorrectly states resistance increases when removing a series resistor always decreases total resistance. Choices C and D wrongly focus on capacitance, which remains unchanged. Safety note: lower discharge resistance in defibrillators delivers higher peak currents to overcome chest impedance.

Question 14

1. Total current decreases because the equivalent resistance increases in parallel.
2. Total current increases because the equivalent resistance decreases in parallel. (correct answer)
3. Total current is unchanged because the resistors have the same values.
4. Total current becomes zero because parallel branches cancel the voltage.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. In Condition 1 (series), Req = R1 + R2 = 1 kΩ + 1 kΩ = 2 kΩ, giving current I = V/R = 9 V / 2 kΩ = 4.5 mA. In Condition 2 (parallel), 1/Req = 1/1 kΩ + 1/1 kΩ = 2/1 kΩ, so Req = 0.5 kΩ, giving current I = 9 V / 0.5 kΩ = 18 mA. The current increases by a factor of 4 when switching from series to parallel configuration. This demonstrates why parallel circuits draw more power - they provide multiple current paths, reducing total resistance. Choice A incorrectly claims resistance increases in parallel. Choice C wrongly suggests equal resistors behave identically regardless of configuration. Key principle: at fixed voltage, lower resistance always means higher current (Ohm's law).

Question 15

1. It is larger than $2\,\mu\text{F}$ because series capacitances add.
2. It equals $4\,\mu\text{F}$ because the voltages divide.
3. It is smaller than $2\,\mu\text{F}$ because series capacitances reduce total capacitance. (correct answer)
4. It is unchanged because both capacitors store the same charge.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this neuron stimulation setup, connecting two identical 2 μF capacitors in series yields an equivalent capacitance of 1 μF, smaller than a single 2 μF. This occurs because series capacitors share charge, effectively reducing total capacitance below the individual values. A distractor like choice A fails by mistakenly applying parallel addition rules to a series configuration. For a transferable check, use the reciprocal sum formula for series capacitors and compare to individual values. Ensure the context specifies series to apply the correct equivalence.

Question 16

1. The time constant doubles because total capacitance increases. (correct answer)
2. The time constant halves because total capacitance decreases.
3. The time constant is unchanged because the resistor sets the charging rate.
4. The time constant doubles because total resistance increases in parallel.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this cell-membrane model, adding a second capacitor in parallel doubles the total capacitance, thus doubling the time constant τ = RC with unchanged resistance. This happens because parallel capacitors increase charge storage capacity, slowing the charging process. A distractor like choice B fails by confusing parallel with series, where capacitance would decrease instead. As a transferable check, compute τ before and after adding components to see effects on transient response. Confirm if components are added in series or parallel to predict τ changes accurately.

Question 17

1. It is $5\,\text{nF}$ because parallel capacitances add. (correct answer)
2. It is $0.8\,\text{nF}$ because parallel capacitances add reciprocals.
3. It is $4\,\text{nF}$ because the larger capacitor dominates.
4. It is $1\,\text{nF}$ because the smaller capacitor limits charge storage.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this sensor filter, connecting 1 nF and 4 nF in parallel gives 5 nF equivalent, summing individual capacitances. This is because parallel capacitors share voltage, adding their charge capacities directly. A distractor like choice B fails by applying the series reciprocal rule to a parallel setup. As a transferable check, add capacitances for parallel and use reciprocals for series to find equivalents. Always note the connection type to apply the proper formula.

Question 18

1. It is less than $1\,\text{k}\Omega$ because a parallel branch always lowers total resistance below any series element.
2. It equals $5\,\text{k}\Omega$ because all resistors add in series.
3. It is greater than $1\,\text{k}\Omega$ because the parallel pair has finite resistance added in series with $R_1$. (correct answer)
4. It is exactly $2\,\text{k}\Omega$ because two equal resistors in parallel equal their sum.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this oximeter design, the total resistance is 1 kΩ plus the 1 kΩ parallel pair, exceeding 1 kΩ alone. This results from adding the finite parallel resistance in series, increasing the overall value. A distractor like choice A fails by wrongly claiming parallel always drops below any series element, ignoring the series addition. For a transferable check, break down mixed circuits into series and parallel subsections. Compute step-by-step to compare with single components.

Question 19

1. It increases because the larger resistor draws more current.
2. It decreases because the total series resistance increases. (correct answer)
3. It is unchanged because the battery voltage is constant.
4. It decreases because capacitance increases in series.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this EEG device, increasing R2 raises total series resistance, decreasing battery current. This is because series sums resistances, reducing overall current per Ohm's law. A distractor like choice A fails by misattributing current draw to larger resistors. As a transferable check, recalculate total Req after changes to predict I = V/Req. Confirm series additions increase Req and decrease current.

Question 20

1. Discharge becomes slower because $R_{\text{eq}}$ increases in parallel.
2. Discharge becomes faster because $R_{\text{eq}}$ decreases in parallel. (correct answer)
3. Discharge is unchanged because the capacitor sets the time constant alone.
4. Discharge stops because adding a resistor in parallel prevents current flow.

Explanation: This question tests understanding of resistors and capacitors in series and parallel. Resistors in series add up to increase total resistance, while capacitors in parallel add to increase total capacitance. In this membrane model, adding parallel resistance halves Req, reducing τ and speeding discharge. This is because lower Req allows faster current flow. A distractor like choice A fails by claiming Req increases in parallel. As a transferable check, evaluate τ = Req C after parallel additions. Remember parallel resistors decrease Req, accelerating transients.