Resistors and Capacitors in Series and Parallel (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Resistors and Capacitors in Series and Parallel (4C)
A biosensor readout circuit uses a capacitor network to set an integration window. Initially, a single capacitor $C = 20\ \text{nF}$ is used. The designer replaces it with two capacitors in series: $C_1 = 10\ \text{nF}$ and $C_2 = 10\ \text{nF}$. Which outcome would be expected for the equivalent capacitance of the network?
It becomes $10\ \text{nF}$ because equal capacitors in series double the capacitance.
It becomes $20\ \text{nF}$ because the total is conserved when components are split.
It becomes $5\ \text{nF}$ because equal capacitors in series halve the capacitance.
It becomes larger than $20\ \text{nF}$ because series capacitors add directly.
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.
A system response test uses an RC low-pass stage where the resistor is formed by two series resistors ($R_1 = 10\ \text{k}\Omega$, $R_2 = 10\ \text{k}\Omega$) feeding a capacitor to ground ($C = 1\ \mu\text{F}$). Without changing $C$, the engineer shorts (bypasses) $R_2$ so only $R_1$ remains in series with the capacitor. Based on the configuration, which change is most likely to occur in the charging rate of the capacitor following a step input?
Charging becomes faster because the equivalent capacitance increases when a resistor is removed.
Charging is unchanged because resistors only affect steady-state current, not transients.
Charging becomes faster because the series resistance decreases.
Charging becomes slower because the capacitor now has less voltage across it.
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.
A microfluidic sensor uses a voltage divider with two resistors in series ($R_1=1\ \text{k}\Omega$, $R_2=9\ \text{k}\Omega$) powered by $V=5\ \text{V}$. A small capacitor ($C=100\ \text{nF}$) is placed in parallel with $R_2$ to filter noise. Which statement best describes the behavior of the circuit immediately after a step increase in the supply voltage?
The capacitor initially behaves like a short circuit, pulling the divider output toward the lower node before settling.
The capacitor decreases the total series resistance permanently, changing the final divider ratio.
The capacitor increases the series resistance, slowing the current through both resistors equally.
The capacitor initially behaves like an open circuit, so the divider output jumps to its final value instantly.
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.
An ECG front-end includes two capacitors ($C_1=1\ \mu\text{F}$ and $C_2=1\ \mu\text{F}$) placed in parallel across an electrode interface to increase charge storage. Which outcome would be expected when a component is added in parallel?
Total resistance increases, so less charge is stored at the same voltage.
Total resistance decreases, so the voltage across each capacitor must decrease.
Total capacitance increases, allowing more charge to be stored at the same voltage.
Total capacitance decreases, allowing less charge to be stored at the same voltage.
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.
In an experiment modeling myelinated axons, two membrane segments are represented by capacitors $C_1$ and $C_2$ in series (each $=5\ \text{pF}$). The goal is to reduce effective capacitance to speed voltage changes. Which statement best describes the behavior of the circuit?
Series capacitors yield a smaller effective capacitance than either capacitor alone.
Effective capacitance is unchanged because capacitance depends only on voltage.
Series capacitors behave like series resistors, so effective capacitance is the sum.
Series capacitors yield a larger effective capacitance than either capacitor alone.
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.
A biomedical device uses a battery and two branches in parallel. Branch 1 is a resistor $R=1\ \text{k}\Omega$. Branch 2 is a capacitor $C=10\ \mu\text{F}$ in series with a resistor $R=1\ \text{k}\Omega$. Which statement best describes the current in Branch 2 long after the battery is connected (DC steady state)?
Branch 2 current increases over time because the capacitor gradually becomes a better conductor.
Branch 2 current is nonzero because the series resistor allows DC through the capacitor.
Branch 2 current goes to zero because the capacitor blocks DC at steady state.
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.
A model of a cell membrane uses a capacitor $C_m$ in parallel with a membrane resistance $R_m$. If a drug opens additional leak channels, $R_m$ decreases. Which statement best describes the effect on the membrane’s ability to maintain a voltage difference after a brief current pulse (qualitative)?
Voltage decays faster because the effective capacitance decreases in parallel.
Voltage is unaffected because capacitors determine only steady-state behavior.
Voltage decays faster because the effective resistance is lower, increasing leakage.
Voltage decays slower because lower resistance reduces discharge current.
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.
A student builds a circuit with two capacitors in parallel connected to a 9 V battery. They then disconnect the battery and connect the capacitor pair across a resistor $R$. Which statement best describes the initial voltage across the resistor right after connection (assuming ideal wires)?
It is 9 V because the capacitors in parallel share the same voltage as the battery they were charged with.
It is 18 V because the two capacitor voltages add in parallel.
It is 0 V because capacitors cannot provide current to a resistor.
It is 4.5 V because the voltage splits between the two capacitors in parallel.
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.
In a patch-clamp amplifier used to record a neuronal membrane, an input protection network includes two resistors $R_1=2\ \text{M}\Omega$ and $R_2=2\ \text{M}\Omega$ placed in parallel between the electrode lead and ground. The electrode sees an applied step of $V=20\ \text{mV}$ relative to ground. Which statement best describes the behavior of the circuit at the electrode lead regarding the equivalent resistance of the protection network?
The equivalent resistance is $0.5\ \text{M}\Omega$ because resistances in parallel halve twice for two components.
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.
The equivalent resistance is $2\ \text{M}\Omega$ because identical resistors in parallel behave like a single resistor of the same value.
The equivalent resistance is $4\ \text{M}\Omega$ because parallel resistances add directly.
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.
A lab team models a cell membrane as a capacitor and tests how rearranging capacitors changes charge storage at fixed voltage. Two identical capacitors, $C_1=1\ \mu\text{F}$ and $C_2=1\ \mu\text{F}$, are connected in series across a $V=10\ \text{V}$ source. Which outcome would be expected for the equivalent capacitance compared with a single $1\ \mu\text{F}$ capacitor across the same source?
The equivalent capacitance equals $1\ \mu\text{F}$ because series connection does not change capacitance for identical components.
The equivalent capacitance is smaller than $1\ \mu\text{F}$, so less total charge is stored for the same applied voltage.
The equivalent capacitance equals $2\ \mu\text{F}$ because capacitances in series add directly.
The equivalent capacitance is larger than $1\ \mu\text{F}$, so more total charge is stored for the same applied voltage.
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.