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 decreases from $25\ \text{mA}$ to $20\ \text{mA}$
It changes only if capacitance changes, not resistance
It remains $25\ \text{mA}$ because voltage is unchanged
It increases because resistance and current are directly proportional
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 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?
$0.67\ \mu\text{A}$ because $I=R/V$
$3.0\ \mu\text{A}$
$1.5\ \mu\text{A}$
$6.0\ \mu\text{A}$
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.
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 doubles to $8.0\ \text{mA}$
Current becomes $1.0\ \mA$ because $V=IR^2$
Current halves to $2.0\ \text{mA}$
Current stays $4.0\ \text{mA}$ because voltage sets current
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.
A saline bath is connected to a $9.0\ \text{V}$ source. Two identical electrodes and the bath together are modeled as a single resistor. When $R=3.0\ \text{k}\Omega$, the current is measured. Later, protein buildup increases resistance to $4.5\ \text{k}\Omega$ while voltage stays constant. Which statement about the current is true?
It becomes negative because resistance increased
It is unchanged because the source fixes current
It increases by a factor of 1.5
It decreases by a factor of 1.5
Explanation
This question evaluates circuit elements and Ohm’s Law in resistive baths with changing resistance. Ohm’s Law (V = IR) implies current decreases as resistance increases for fixed voltage. Initial R = 3.0 kΩ at 9.0 V, new R = 4.5 kΩ means I decreases by factor of 3/4.5 = 2/3, or 1 / 1.5, as in choice A. Choice B wrongly suggests current increases, ignoring inverse relation. Check by calculating initial I = V/R, then new I = V/(1.5R) and finding the ratio. This verifies resistance effects on current.
A researcher compares two identical microelectrodes except for coating. Under a fixed $0.10\ \text{V}$ bias, electrode X produces $10\ \mu\text{A}$ and electrode Y produces $5\ \mu\text{A}$ (assume ohmic behavior). Which statement is most consistent with Ohm’s Law?
Their resistances must be equal because the voltage is the same
Electrode Y has half the resistance of electrode X
Electrode Y has twice the resistance of electrode X
Electrode X has higher resistance because its current is larger
Explanation
This question evaluates circuit elements and Ohm’s Law comparing resistances from current measurements. Ohm’s Law (V=IR) implies R = V/I, so lower current means higher resistance at fixed V. Electrode Y has half the current of X, so twice the resistance, as in choice B. Choice D assumes equal resistances from same V, ignoring current differences. Verify by calculating R_X = V / I_X and R_Y = V / I_Y, then ratio R_Y / R_X = I_X / I_Y. This quantifies resistance differences.
In a tissue-engineering lab, a saline-filled microchannel is modeled as a resistor of $R=2.0\ \text{k}\Omega$ connected to a constant-voltage source of $V=6.0\ \text{V}$. If the channel’s ionic concentration is reduced such that its resistance increases to $4.0\ \text{k}\Omega$ while the source voltage remains fixed, which outcome would be most consistent with Ohm’s Law for the current through the channel?
The current is unchanged because voltage is constant
The current increases by a factor of 2
The current decreases because the voltage must drop when resistance increases
The current decreases by a factor of 2
Explanation
This question tests understanding of how current changes when resistance changes in a circuit with constant voltage, a direct application of Ohm's Law. Ohm's Law states that V = IR, which can be rearranged to I = V/R, showing that current is inversely proportional to resistance when voltage is constant. In this microchannel system, the initial current is I₁ = 6.0 V / 2.0 kΩ = 3.0 mA, and when resistance doubles to 4.0 kΩ, the new current becomes I₂ = 6.0 V / 4.0 kΩ = 1.5 mA. Since 1.5 mA is half of 3.0 mA, the current decreases by a factor of 2, making choice A correct. Choice C incorrectly assumes current remains constant when resistance changes, failing to recognize the inverse relationship between current and resistance. To verify understanding, always check that when resistance doubles in a constant-voltage circuit, current must halve, and vice versa.
In an electrophysiology rig, a membrane patch is approximated as a resistor. When the clamp applies $V=80\ \text{mV}$, the measured current is $I=4.0\ \text{nA}$. If the same patch is later measured at $V=120\ \text{mV}$ and behaves ohmically over this range, which current is most consistent with Ohm’s Law?
$I=1.5\ \text{nA}$
$I=160\ \text{nA}$
$I=6.0\ \text{nA}$
$I=2.7\ \text{nA}$
Explanation
This question tests the linear relationship in Ohm's Law, requiring students to recognize that resistance remains constant for an ohmic device. From the initial conditions, we can calculate the membrane resistance: R = V/I = 80 mV / 4.0 nA = 20 MΩ. Since the patch behaves ohmically (linearly), this resistance remains constant at the new voltage. At 120 mV, the current becomes I = V/R = 120 mV / 20 MΩ = 6.0 nA, making choice A correct. Choice B incorrectly calculates a different resistance value, failing to recognize that ohmic behavior means constant resistance. To verify ohmic behavior, check that the V/I ratio remains constant across different voltage levels - this ratio is the resistance.
During a benchtop test of a microfluidic conductivity sensor, the device is modeled as a single resistor. With $V=5.0\ \text{V}$ applied, the measured current is $I=2.0\ \text{mA}$. A technician reports that the resistance is $2.5\ \text{k}\Omega$. Which statement is most consistent with Ohm’s Law?
Resistance cannot be determined without capacitance data
The resistance should be $2.5\ \text{k}\Omega$ because $R=V/I$
The resistance should be $10\ \text{k}\Omega$, not $2.5\ \text{k}\Omega$
The resistance should be $0.40\ \text{k}\Omega$ because $R=I/V$
Explanation
This question tests direct application of Ohm's Law to verify a resistance calculation. Ohm's Law states V = IR, which rearranges to R = V/I for finding resistance from voltage and current measurements. With V = 5.0 V and I = 2.0 mA = 0.002 A, the resistance is R = 5.0 V / 0.002 A = 2500 Ω = 2.5 kΩ, confirming the technician's report in choice B. Choice C incorrectly inverts the formula to R = I/V, a common algebraic error that yields nonsensical units (A/V instead of V/A = Ω). To verify resistance calculations, always check that units work out correctly: volts divided by amperes gives ohms, and the numerical result should make physical sense for the application.
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?
$V_{R_2}=0.80\ \text{V}$
$V_{R_2}=2.4\ \text{V}$ because voltages add across a resistor
$V_{R_2}=0.40\ \text{V}$ because larger resistance gets smaller voltage
$V_{R_2}=1.2\ \text{V}$ because each series element has the same 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.
In a conductivity measurement, a solution is modeled as an ohmic resistor. At $V=0.50\ \text{V}$, the current is $2.5\ \text{mA}$. If the same solution is cooled such that its resistance increases by $20%$ while the applied voltage is held constant, what happens to the current?
It stays at $2.5\ \text{mA}$ because voltage is fixed
It decreases to about $2.1\ \text{mA}$
It changes sign because resistance increased
It increases to about $3.0\ \text{mA}$
Explanation
This question evaluates circuit elements and Ohm’s Law with percentage resistance change. Ohm’s Law I = V / R; 20% R increase means new R=1.2 * original, new I = original I / 1.2 ≈ 2.5mA / 1.2 ≈ 2.1mA, matching choice A. Choice B wrongly increases I, ignoring inverse. To check, compute original R= V/I=0.50V/2.5mA=200Ω, new R=240Ω, I=0.50/240≈0.00208A=2.1mA. This verifies percentage effects.