MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4c Circuit Elements Ohms Law

What this quiz covers

This quiz focuses on 4c Circuit Elements Ohms Law, 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 increases because resistance and current are directly proportional
2. It decreases from $25\ \text{mA}$ to $20\ \text{mA}$ (correct answer)
3. It remains $25\ \text{mA}$ because voltage is unchanged
4. It changes only if capacitance changes, not resistance

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.

Question 2

1. $6.0\ \mu\text{A}$
2. $3.0\ \mu\text{A}$
3. $1.5\ \mu\text{A}$ (correct answer)
4. $0.67\ \mu\text{A}$ because $I=R/V$

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.

Question 3

1. Current halves to $2.0\ \text{mA}$
2. Current doubles to $8.0\ \text{mA}$ (correct answer)
3. Current stays $4.0\ \text{mA}$ because voltage sets current
4. Current becomes $1.0\ \mA$ because $V=IR^2$

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.

Question 4

1. It decreases by a factor of 1.5 (correct answer)
2. It increases by a factor of 1.5
3. It is unchanged because the source fixes current
4. It becomes negative because resistance increased

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.

Question 5

1. Electrode Y has half the resistance of electrode X
2. Electrode Y has twice the resistance of electrode X (correct answer)
3. Electrode X has higher resistance because its current is larger
4. Their resistances must be equal because the voltage is the same

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.

Question 6

1. The current decreases by a factor of 2 (correct answer)
2. The current increases by a factor of 2
3. The current is unchanged because voltage is constant
4. The current decreases because the voltage must drop when resistance increases

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.

Question 7

1. $I=6.0\ \text{nA}$ (correct answer)
2. $I=2.7\ \text{nA}$
3. $I=1.5\ \text{nA}$
4. $I=160\ \text{nA}$

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.

Question 8

1. The resistance should be $10\ \text{k}\Omega$, not $2.5\ \text{k}\Omega$
2. The resistance should be $2.5\ \text{k}\Omega$ because $R=V/I$ (correct answer)
3. The resistance should be $0.40\ \text{k}\Omega$ because $R=I/V$
4. Resistance cannot be determined without capacitance data

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.

Question 9

1. It decreases to about $2.1\ \text{mA}$ (correct answer)
2. It increases to about $3.0\ \text{mA}$
3. It stays at $2.5\ \text{mA}$ because voltage is fixed
4. It changes sign because resistance increased

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.

Question 10

1. $9.5\ \text{V}$ (correct answer)
2. $0.5\ \text{V}$
3. $10\ \text{V}$ because series elements each get the full voltage
4. $95\ \text{V}$ because resistance multiplies voltage

Explanation: This question probes circuit elements and Ohm's Law in series voltage division for nerve electrodes. Ohm's Law (V = IR) means voltage drops proportionally to resistance in series. Total R = 5Ω + 95Ω = 100Ω, I = 10V / 100Ω = 0.1A, V_95 = I * 95 = 9.5V, confirming choice A. Choice C assumes full voltage per element, misunderstanding series division. Verify by finding I from total V/R_total, then V_component = I * R_component. This ensures comprehension of voltage distribution.

Question 11

1. The sample resistance increased from $2\ \text{k}\Omega$ to $3\ \text{k}\Omega$ (correct answer)
2. The sample resistance decreased because voltage increased
3. The current must have increased because voltage increased
4. Resistance is unchanged; only capacitance can change the voltage

Explanation: This question examines circuit elements and Ohm's Law under constant current in tissue resistance measurement. Ohm's Law (V = IR) shows that for fixed current, voltage is proportional to resistance. The tissue is a resistor; initial V = 2.0 V at I = 1.0 mA gives R = 2 kΩ, new V = 3.0 V implies R = 3 kΩ, so resistance increased, as in choice A. Choice B incorrectly assumes resistance decreases with voltage increase, reversing proportionality. Verify by calculating R = V/I for both cases and comparing. This check highlights resistance changes in constant-current setups.

Question 12

1. $2\ \text{nA}$
2. $0.2\ \text{nA}$ (correct answer)
3. $200\ \text{nA}$
4. $0.2\ \text{mA}$

Explanation: This question tests circuit elements and Ohm's Law in modeling neuronal membrane resistance. Ohm's Law states V = IR, so current I = V / R for a given voltage across a resistor. The neuron membrane is modeled as R_m = 100 MΩ with ΔV = 20 mV. Thus, I = 20 mV / 100 MΩ = 0.2 nA, matching choice B. Choice A errs by inverting the formula, using I = R / V instead. To confirm understanding, rearrange Ohm's Law to solve for I and plug in values, then check units for consistency. This process reinforces application to biological resistances.

Question 13

1. It increases from $0.20\ \text{V}$ to $0.30\ \text{V}$ (correct answer)
2. It decreases from $0.20\ \text{V}$ to $0.13\ \text{V}$
3. It stays at $0.20\ \text{V}$ because current is fixed
4. It becomes $1.7\ \text{V}$ because resistance adds to voltage

Explanation: This question tests understanding of constant-current sources and how voltage responds to resistance changes according to Ohm's Law. With a constant-current source, I remains fixed while V = IR must adjust when R changes. Initially, V₁ = (200 μA)(1.0 kΩ) = 200 mV = 0.20 V. When resistance increases to 1.5 kΩ, the new voltage becomes V₂ = (200 μA)(1.5 kΩ) = 300 mV = 0.30 V, confirming choice A. Choice B incorrectly assumes voltage decreases when resistance increases, contradicting the direct proportionality in V = IR when current is constant. To verify constant-current circuit behavior, remember that voltage and resistance are directly proportional - if one increases by 50%, the other must also increase by 50%.

Question 14

1. Resistor voltage drop increases, so current increases (correct answer)
2. Resistor voltage drop decreases, so current increases
3. Resistor voltage drop increases, so current decreases
4. Current stays constant because resistance fixes voltage

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.

Question 15

1. $0.16\ \text{V}$
2. $1.0\ \text{V}$ (correct answer)
3. $0.40\ \text{V}$ because voltage is a property of the material
4. $2.0\ \text{V}$ because voltage increases with $I^2$

Explanation: This question assesses circuit elements and Ohm's Law with varying current in resistive strips. Ohm's Law, V = IR, shows voltage proportional to current for fixed resistance. Initial V = 0.40V at 2.0mA gives R = 200Ω, at 5.0mA V = 5.0mA * 200Ω = 1.0V, matching choice B. Choice D incorrectly uses V ~ I^2, confusing with power. To check, calculate R from initial, then V_new = I_new * R. This verifies direct proportionality.

Question 16

1. $0.20\ \text{A}$ (correct answer)
2. $0.50\ \text{A}$
3. $2.0\ \text{A}$
4. $10\ \text{A}$ because resistances add to increase current

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.

Question 17

1. Current decreases to about $42\ \text{mA}$ (correct answer)
2. Current increases to $60\ \text{mA}$
3. Current stays at $50\ \text{mA}$ because voltage is unchanged
4. Current becomes $1.0\ \text{A}$ because resistance increases

Explanation: This question tests understanding of how current changes when resistance increases in a constant-voltage circuit. Initially, the circuit resistance is R₁ = V/I = 5.0 V / 50 mA = 100 Ω. When resistance increases by 20%, the new resistance becomes R₂ = 1.20 × 100 Ω = 120 Ω. With voltage constant at 5.0 V, the new current is I₂ = V/R₂ = 5.0 V / 120 Ω = 0.0417 A ≈ 42 mA, confirming choice A. Choice C incorrectly assumes current remains constant when resistance changes, failing to apply the inverse relationship between current and resistance in Ohm's Law. To verify such problems, remember that a 20% increase in resistance causes approximately a 17% decrease in current (from 50 mA to 42 mA), not a 20% decrease, due to the inverse relationship.

Question 18

1. The current through $R_d$ is larger because it has higher resistance
2. The current through $R_s$ is $0.20\ \text{mA}$ (correct answer)
3. The current through $R_s$ is $0.05\ \text{mA}$ because currents split equally in parallel
4. 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.

Question 19

1. $12\ \mu\text{A}$
2. $6.0\ \mu\text{A}$
3. $3.0\ \mu\text{A}$ (correct answer)
4. $0.60\ \mu\text{A}$ because microamps scale with millivolts

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.

Question 20

1. Current becomes $12\ \text{mA}$ (correct answer)
2. Current becomes $33\ \text{mA}$
3. Current remains $20\ \text{mA}$ because resistance is constant
4. Current becomes $1.2\ \text{A}$ because volts convert directly to amps

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.