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 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 decreases the total series resistance permanently, changing the final divider ratio.
The capacitor initially behaves like a short circuit, pulling the divider output toward the lower node before settling.
The capacitor initially behaves like an open circuit, so the divider output jumps to its final value instantly.
The capacitor increases the series resistance, slowing the current through both resistors equally.
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 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.
Total resistance increases, so less charge is 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.
A researcher builds a stimulator where a capacitor $C=10\ \mu\text{F}$ is placed in series with a tissue load modeled as a resistor $R=1\ \text{k}\Omega$. The capacitor is used to block DC while allowing transient pulses. Based on the configuration, which change is most likely to occur if a second identical capacitor is added in series with the first?
The effective resistance decreases, causing a shorter transient response.
The effective capacitance decreases, causing a shorter transient response.
The effective resistance increases, causing a longer transient response.
The effective capacitance increases, causing a longer transient response.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. Capacitors in series combine like resistors in parallel: 1/C_total = 1/C1 + 1/C2, so adding a second identical 10 μF capacitor in series results in C_total = 5 μF, which is half the original capacitance. The RC time constant τ = RC determines the transient response duration, so halving the capacitance halves the time constant, resulting in a shorter transient response. The correct answer identifies that effective capacitance decreases in series configuration. Answer A incorrectly states capacitance increases, while answers C and D incorrectly focus on resistance changes when only capacitance is being modified.
In a patch-clamp amplifier input stage, two resistors ($R_1=2\ \text{M}\Omega$ and $R_2=2\ \text{M}\Omega$) are placed in parallel to reduce thermal noise while maintaining a high input resistance. A membrane capacitance $C_m=20\ \text{pF}$ is in parallel with the input. Which statement best describes the behavior of the circuit at steady DC after the input has been held at a constant voltage for a long time?
The parallel resistors do not change total resistance, and the capacitor acts like a short circuit at steady state.
The parallel resistors increase total resistance, and the capacitor continues to pass DC current.
The parallel resistors decrease total resistance, and the capacitor carries no DC current at steady state.
The parallel resistors decrease total capacitance, and the capacitor discharges instantly at steady state.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. Resistors in parallel have a combined resistance given by 1/R_total = 1/R1 + 1/R2, so two 2 MΩ resistors in parallel give 1 MΩ total resistance, which is less than either individual resistor. At steady DC state, capacitors act as open circuits because they block DC current after fully charging. The correct answer recognizes that parallel resistors decrease total resistance and capacitors carry no DC current at steady state. Answer A incorrectly states that parallel resistors increase resistance, while answers C and D make false claims about capacitor behavior at steady state.
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.
Series capacitors yield a larger 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.
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 wearable sensor has two resistive strain gauges that must both carry the same current; they are wired in series ($R_1=300\ \Omega$, $R_2=700\ \Omega$). Based on the configuration, which change is most likely to occur if the gauges are rewired in parallel across the same supply voltage?
Total resistance decreases, so voltage across each gauge decreases below the supply.
Total resistance decreases, so total current drawn from the supply increases.
Total resistance increases, so voltage across each gauge increases above the supply.
Total resistance increases, so total current drawn from the supply decreases.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. When resistors are in series, the total resistance is R1 + R2 = 300 + 700 = 1000 Ω, but when rewired in parallel, the total resistance becomes (1/300 + 1/700)^-1 ≈ 210 Ω. Since V = IR, with the same supply voltage and lower total resistance in parallel configuration, the total current drawn from the supply increases. The correct answer recognizes that parallel resistors decrease total resistance and increase current draw. Answer A incorrectly states resistance increases, while answers C and D make incorrect claims about voltage behavior.
A lab uses an RC timing circuit to gate a drug-delivery microvalve. The resistor network is two identical resistors ($R$ and $R$) in parallel feeding a capacitor $C$ to ground (a standard low-pass). Which statement best describes the behavior of the circuit compared with using a single resistor $R$ (all else unchanged)?
The time constant decreases because the effective capacitance is smaller.
The time constant decreases because the effective resistance is smaller.
The time constant increases because resistors in parallel add.
The time constant is unchanged because capacitance dominates the transient.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. Two identical resistors R in parallel have an effective resistance of R/2, which is smaller than a single resistor R. The RC time constant τ = RC determines the circuit's transient response time, so halving the resistance halves the time constant, making the transient response faster. The correct answer identifies that parallel resistors decrease effective resistance and time constant. Answer A incorrectly claims resistors in parallel add, answer C incorrectly states the time constant is unchanged, and answer D incorrectly focuses on capacitance changes.
A defibrillator training model uses two capacitors ($C_1=50\ \mu\text{F}$, $C_2=50\ \mu\text{F}$) in parallel to store energy at $V=100\ \text{V}$. Which statement best describes the behavior of the circuit relative to using only one of the capacitors at the same voltage?
Total capacitance is halved, so stored charge is halved.
Total resistance doubles, so stored charge doubles.
Total capacitance doubles, so stored charge doubles.
Total capacitance doubles, so the voltage across each capacitor doubles.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. Capacitors in parallel add directly: C_total = C1 + C2 = 50 + 50 = 100 μF, which is double the capacitance of a single 50 μF capacitor. Since Q = CV, doubling the capacitance at the same voltage (100V) doubles the stored charge and energy storage capability. The correct answer recognizes that parallel capacitors double the total capacitance and stored charge. Answer A incorrectly states capacitance is halved, answer C incorrectly invokes resistance, and answer D makes an incorrect claim about voltage distribution.
A student models two ion channels as resistors in parallel ($R_1=5\ \text{M}\Omega$, $R_2=5\ \text{M}\Omega$) connected across the same membrane potential. Which outcome would be expected when a component is added in parallel (i.e., opening an additional identical channel)?
Voltage across each channel decreases because current splits.
Total resistance increases, reducing total current at the same voltage.
Total capacitance increases, increasing total current at the same voltage.
Total resistance decreases, increasing total current at the same voltage.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. When resistors are in parallel, the total resistance decreases according to 1/R_total = 1/R1 + 1/R2. Adding another identical 5 MΩ channel in parallel further reduces the total resistance. By Ohm's law (I = V/R), with the same membrane potential and lower total resistance, the total current increases. The correct answer recognizes that parallel resistors decrease resistance and increase current. Answer A incorrectly states resistance increases, answer C incorrectly invokes capacitance, and answer D makes an incorrect claim about voltage distribution.
A stimulation circuit places two resistors in series ($R_1=1\ \text{k}\Omega$, $R_2=1\ \text{k}\Omega$) with a capacitor $C=1\ \mu\text{F}$ connected in parallel with $R_2$. The output is measured across $R_2$ (and the capacitor). Which statement best describes the circuit’s DC behavior long after a constant voltage is applied?
The capacitor behaves as an open circuit, so the output is set by the resistor divider $R_1$ and $R_2$.
The capacitor behaves as a short circuit, so the output equals the full supply voltage.
The capacitor increases the effective resistance of $R_2$, raising the output above the divider value.
The capacitor behaves as a short circuit, so the output is forced to 0 V.
Explanation
This question tests understanding of resistors and capacitors in series and parallel. At DC steady state (long after a constant voltage is applied), capacitors act as open circuits, carrying no current. With the capacitor acting as an open circuit, the output voltage is determined solely by the resistive voltage divider formed by R1 and R2, giving Vout = Vin × R2/(R1+R2) = Vin/2 for equal resistors. The correct answer recognizes that capacitors are open circuits at DC steady state. Answer B and C incorrectly describe the capacitor as a short circuit, while answer D makes an incorrect claim about resistance changes.