Circuit Elements and Ohm’s Law (4C)
Help Questions
MCAT Chemical and Physical Foundations of Biological Systems › Circuit Elements and Ohm’s Law (4C)
A wearable biosensor uses a resistive strain gauge ($R=200\ \Omega$) powered by a $5.0\ \text{V}$ supply. During motion, the gauge resistance increases to $250\ \Omega$ while the supply voltage remains constant. How would the current through the gauge change, consistent with Ohm’s Law?
It increases because resistance and current are directly proportional
It remains $25\ \text{mA}$ because voltage is unchanged
It changes only if capacitance changes, not resistance
It decreases from $25\ \text{mA}$ to $20\ \text{mA}$
Explanation
This question evaluates application of circuit elements and Ohm’s Law to variable resistance in biosensors. Ohm’s Law (V = IR) indicates that for constant voltage, current decreases as resistance increases. The strain gauge resistor changes from 200 Ω to 250 Ω under constant 5.0 V supply. Initial I = 5 V / 200 Ω = 25 mA, new I = 5 V / 250 Ω = 20 mA, so it decreases, aligning with choice B. Choice C errs by assuming current unchanged with voltage, ignoring resistance's role in Ohm’s Law. For verification, use I = V/R for both values and compare, highlighting the inverse proportionality. This approach aids in predicting current changes in variable-resistance sensors.
A handheld glucometer applies $0.30\ \text{V}$ across an enzyme-coated test strip that behaves ohmically at the measurement point. The measured current is $6.0\ \mu\text{A}$. If a different strip has twice the resistance under the same voltage, what current is expected?
$6.0\ \mu\text{A}$
$3.0\ \mu\text{A}$
$0.60\ \mu\text{A}$ because microamps scale with millivolts
$12\ \mu\text{A}$
Explanation
This question assesses circuit elements and Ohm’s Law with varying resistance in test strips. Ohm’s Law, V = IR, means current inversely proportional to resistance at constant voltage. Initial I = 6.0 μA at 0.30 V, so doubling R halves I to 3.0 μA, supporting choice C. Choice D misapplies I = R/V, inverting the relationship. To verify, compute initial R = V/I, double it, then find new I = V / (2R). This method ensures understanding of inverse proportionality in fixed-voltage circuits.
A simple model of skin impedance uses two resistors in parallel: dry pathway $R_d=100\ \text{k}\Omega$ and sweat pathway $R_s=25\ \text{k}\Omega$. A $5.0\ \text{V}$ measurement voltage is applied across both. Which statement about branch currents is true?
The current through $R_s$ is $0.20\ \text{mA}$
The current through $R_s$ is $0.05\ \text{mA}$ because currents split equally in parallel
The current through $R_d$ is larger because it has higher resistance
The current through $R_s$ depends on capacitance, not resistance
Explanation
This question assesses circuit elements and Ohm’s Law in parallel skin impedance models. Ohm’s Law applies to branches; current splits inversely with resistance, I_branch = V / R_branch. For R_s=25kΩ, I_s=5.0V / 25kΩ=0.20mA, matching choice B. Choice C assumes equal split, ignoring resistance differences. Verify by calculating each I = V/R, noting lower R gets higher I. This checks parallel current division.
Two resistive heating elements in a temperature-controlled incubator are modeled as resistors in series: $R_1=20\ \Omega$ and $R_2=30\ \Omega$. A $10\ \text{V}$ supply is applied across the series pair. Based on Ohm’s Law, what is the current through the circuit?
$10\ \text{A}$ because resistances add to increase current
$0.50\ \text{A}$
$2.0\ \text{A}$
$0.20\ \text{A}$
Explanation
This question tests circuit elements and Ohm’s Law for series resistors in heating elements. Ohm’s Law gives I = V / R_total, with R_total = R1 + R2 in series. Here, R_total = 20Ω + 30Ω = 50Ω, I = 10V / 50Ω = 0.20A, confirming choice A. Choice D wrongly adds resistances to increase current, reversing effect. To verify, sum series resistances and divide V by total. This method applies to any series circuit current calculation.
A researcher applies $1.2\ \text{V}$ across a series pair of resistors modeling two tissue layers: $R_1=400\ \Omega$ and $R_2=800\ \Omega$. Which statement about the voltage across $R_2$ is true?
$V_{R_2}=1.2\ \text{V}$ because each series element has the same voltage
$V_{R_2}=0.80\ \text{V}$
$V_{R_2}=2.4\ \text{V}$ because voltages add across a resistor
$V_{R_2}=0.40\ \text{V}$ because larger resistance gets smaller voltage
Explanation
This question tests circuit elements and Ohm’s Law in series voltage drops for tissue layers. In series, V drops proportionally: V_R2 = V_total * (R2 / (R1+R2)) = 1.2V * (800/(400+800)) = 1.2 * (800/1200) = 1.2 * (2/3) = 0.80V, confirming choice B. Choice A incorrectly assigns smaller voltage to larger R, reversing proportionality. Verify by finding I = V_total / R_total, then V_R2 = I * R2. This ensures accurate voltage division calculation.
A researcher uses a $3.0\ \text{V}$ coin cell to power a biosensor. The sensor’s input stage is approximated as a single resistor. When the resistance is $1.0\ \text{M}\Omega$, the current is $3.0\ \mu\text{A}$. If a firmware change increases the effective resistance to $2.0\ \text{M}\Omega$ at the same voltage, which current is expected?
$3.0\ \mu\text{A}$
$1.5\ \mu\text{A}$
$0.67\ \mu\text{A}$ because $I=R/V$
$6.0\ \mu\text{A}$
Explanation
This question evaluates circuit elements and Ohm’s Law with resistance changes in biosensors. Ohm’s Law (V = IR) implies current halves if resistance doubles at constant voltage. Initial I = 3.0 μA at 3.0V with 1.0 MΩ, new R=2.0 MΩ gives I=1.5 μA, supporting choice C. Choice D misuses I=R/V, which is incorrect. Verify by computing I = V/R for both resistances. This reinforces inverse relationship in constant-voltage setups.
A pulse oximeter’s LED driver is simplified as a $50\ \Omega$ resistor in series with the LED, and during a certain operating point the LED drop is treated as constant. If the supply voltage is increased slightly and the series resistance is unchanged, which change is most consistent with Ohm’s Law for the resistor’s voltage drop and current?
Resistor voltage drop increases, so current decreases
Current stays constant because resistance fixes voltage
Resistor voltage drop increases, so current increases
Resistor voltage drop decreases, so current increases
Explanation
This question tests circuit elements and Ohm’s Law in series with constant LED drop. Ohm’s Law for the resistor: V_resistor = I * R, where V_resistor = V_supply - V_LED. Increasing V_supply increases V_resistor (assuming V_LED constant), thus increases I = V_resistor / R, matching choice A. Choice B incorrectly decreases V_resistor, misunderstanding supply increase. To check, model as voltage divider with fixed V_LED, compute I from (V_supply - V_LED)/R. This verifies current response to voltage changes.
In an electrophoresis setup, the buffer between electrodes is approximated as an ohmic resistor. At $V=100\ \text{V}$, the current is $20\ \text{mA}$. If the experimenter lowers the applied voltage to $60\ \text{V}$ without changing the buffer, which outcome is most consistent with Ohm’s Law?
Current remains $20\ \text{mA}$ because resistance is constant
Current becomes $12\ \text{mA}$
Current becomes $1.2\ \text{A}$ because volts convert directly to amps
Current becomes $33\ \text{mA}$
Explanation
This question probes understanding of circuit elements and Ohm’s Law in electrophoresis setups with constant resistance. Ohm’s Law, V = IR, implies current is directly proportional to voltage for fixed resistance. The buffer acts as a resistor; initial I = 20 mA at 100 V, so R = 100 V / 20 mA = 5 kΩ. At 60 V, new I = 60 V / 5 kΩ = 12 mA, confirming choice A. Choice C mistakenly holds current constant, confusing fixed voltage with fixed current. To verify, find R from initial conditions then apply new V to get I, ensuring consistency. This transferable method checks proportionality in ohmic systems.
In a microfluidic assay, a saline-filled channel between two Ag/AgCl electrodes behaves as an ohmic resistor. With a constant applied voltage of $1.5\ \text{V}$, the measured current is $0.50\ \text{mA}$. If the channel length is doubled (same cross-sectional area and solution), which outcome is most consistent with Ohm’s Law for the new steady current?
It doubles to $1.0\ \text{mA}$ because the voltage is unchanged
It drops to $0.25\ \text{mA}$ because the resistance increases
It stays at $0.50\ \text{mA}$ because the solution is the same
It increases because longer channels have higher capacitance
Explanation
This question tests understanding of circuit elements and Ohm’s Law in the context of resistive channels in microfluidic systems. Ohm’s Law states that voltage V equals current I times resistance R (V = IR), implying that for a fixed voltage, current is inversely proportional to resistance. In this scenario, the saline-filled channel acts as a resistor where resistance depends on length, with R proportional to length for constant cross-section and resistivity. Doubling the channel length doubles the resistance, so with voltage unchanged, the current halves from 0.50 mA to 0.25 mA, making choice C correct. Choice A fails because it assumes current increases with unchanged voltage, misconstruing the inverse relationship in Ohm’s Law. To verify understanding, calculate the initial resistance as V/I = 1.5 V / 0.50 mA = 3 kΩ, then double it to 6 kΩ and find new I = 1.5 V / 6 kΩ = 0.25 mA. This check reinforces how changes in physical dimensions affect circuit behavior through resistance.
An ion-selective electrode system is approximated as an ohmic resistor during calibration. With $V=2.0\ \text{V}$, the current is $4.0\ \text{mA}$. If the resistance is reduced by half (same applied voltage), which outcome is most consistent with Ohm’s Law?
Current stays $4.0\ \text{mA}$ because voltage sets current
Current halves to $2.0\ \text{mA}$
Current becomes $1.0\ \mA$ because $V=IR^2$
Current doubles to $8.0\ \text{mA}$
Explanation
This question examines circuit elements and Ohm’s Law with halved resistance in electrode systems. Ohm’s Law (V = IR) means current doubles if resistance halves at constant voltage. Initial I=4.0mA at 2.0V, halving R doubles I to 8.0mA, aligning with choice B. Choice A incorrectly halves current, misunderstanding inverse proportionality. Check by finding initial R=V/I, then I_new=V/(R/2)=2*(V/R). This doubles initial I, verifying the relationship.