Question 1

Power fundamentals: Choose the best definition of electrical power in physics and electronics.

Accepted Answer

Introduction / Context: Power connects energy and time. In circuits it determines heating, sizing of supplies, and efficiency. Knowing the correct definition helps relate formulas like P = V * I and P = I^2 * R back to physical meaning. Given Data / Assumptions: Standard physics definitions. Energy measured in joules (J), time in seconds (s), power in watts (W). Electrical work equals charge moved times potential difference. Concept / Approach: Power is the time rate of doing work or transferring energy. Mathematically: P = dW/dt or P = dE/dt. In steady DC circuits, it reduces to algebraic forms like P = V * I because work per unit charge is V and charge per unit time is I. Step-by-Step Solution: Start from definition: P = dE/dt.For electrical systems: E transferred each second by current at voltage V equals V * I.Relate to other forms: P = I^2 * R or P = V^2 / R derived from Ohm’s law. Verification / Alternative check: Unit analysis: V * A = (J/C) * (C/s) = J/s = W, confirming consistency. Why Other Options Are Wrong: Work: That is energy, not its time rate. Conversion of energy: Too vague; power quantifies the rate of conversion. Joules: A unit of energy, not power. Common Pitfalls: Confusing energy with power; using watt-hours (energy) when watts (power) are intended. Final Answer: the rate at which work is done

Question 2

Series drop with a lamp: An 8 Ω resistor is in series with a lamp. The circuit current is 1 A and the applied supply is 20 V DC. What voltage is available across the lamp?

Accepted Answer

Introduction / Context: Series circuits share the same current, but sources divide their voltage among components according to their resistances. This practical problem shows how a series “dropping resistor” reduces the voltage applied to a lamp—useful for quick design checks and troubleshooting. Given Data / Assumptions: Resistor R = 8 Ω in series with a lamp. Current I = 1 A through the series path. Supply voltage V_s = 20 V DC. Lamp modeled as a load that takes the remaining voltage. Concept / Approach: Use Ohm’s law to find the resistor’s voltage drop, then subtract from the source to find the lamp voltage. In any series loop, V_s = V_R + V_lamp. With known I and R, V_R = I * R. Step-by-Step Solution: Compute resistor drop: V_R = I * R = 1 A * 8 Ω = 8 V.Apply KVL: V_lamp = V_s − V_R = 20 V − 8 V.Result: V_lamp = 12 V. Verification / Alternative check: Power check: P_R = I^2 * R = 1^2 * 8 = 8 W; P_lamp = V_lamp * I = 12 * 1 = 12 W; total 8 W + 12 W = 20 W = V_s * I, confirming consistency. Why Other Options Are Wrong: 4 V or 8 V: 8 V is the resistor drop, not the lamp. 4 V is unsupported by the data. 20 V: Would be true only with no series resistor drop (not our case). Common Pitfalls: Forgetting that the same current flows through both elements; mixing up which voltage belongs to which component. Final Answer: 12 V

Question 3

Ideal source behavior: When a battery is connected to a purely series circuit, the current it delivers depends primarily on which single factor?

Accepted Answer

Introduction / Context: Ohm’s law gives a direct relationship among voltage, current, and resistance in simple DC circuits. Recognizing what sets the current helps with power budgeting, fuse sizing, and verifying whether a design will overload a source. Given Data / Assumptions: Ideal battery providing a fixed DC voltage. All elements are in series (single path for current). No internal resistance or nonlinearities considered for the basic principle. Concept / Approach: For an ideal source of voltage V connected to a series resistance R_total, the current is I = V / R_total. Polarity sets direction, not magnitude. Terms such as “primary/secondary” relate to transformers, not DC sources. “Average resistance” is not an electrical parameter; only the algebraic sum matters in series. Step-by-Step Solution: Compute total resistance: R_total = R1 + R2 + ... + Rn.Apply Ohm’s law: I = V / R_total.Conclude that current depends solely on total resistance for a fixed V. Verification / Alternative check: Measure current while adding another series resistor; current decreases according to I_new = V / (R_total + ΔR), confirming dependence on total resistance. Why Other Options Are Wrong: Primary/secondary difference: Not applicable to simple battery circuits. Polarity connections: Affect direction of current, not magnitude for given R_total. Average resistance: Not an electrical standard; magnitude is set by the true total. Common Pitfalls: Ignoring internal battery resistance in real life; while small, it can limit current. The ideal model still teaches the governing dependency. Final Answer: total resistance

Question 4

Ohm’s law in a series circuit: If the total resistance in a series circuit doubles while the applied source voltage remains constant, how will the circuit current change?

Accepted Answer

Introduction / Context: This question tests core understanding of Ohm’s law and how current behaves in a simple series circuit when total resistance changes. In many troubleshooting and design scenarios, recognizing proportional relationships between current, resistance, and voltage is essential for quick, reliable reasoning. Given Data / Assumptions: Total resistance in a series circuit doubles. The source (supply) voltage is constant. Ideal components with no temperature-induced resistance change during comparison. Single-loop series circuit so one current flows through all elements. Concept / Approach: Ohm’s law states I = V / R. If V is constant and R increases, I must decrease proportionally. Specifically, if R becomes 2R, then I becomes V / (2R) = (1/2) * (V / R). Thus current is halved. This inverse proportionality between current and resistance at fixed voltage is a cornerstone of basic circuit analysis. Step-by-Step Solution: Initial current: I1 = V / R_total.After doubling resistance: R_new = 2 * R_total.New current: I2 = V / R_new = V / (2 * R_total) = (1/2) * (V / R_total).Therefore, I2 = I1 / 2, i.e., the current is halved. Verification / Alternative check: Use numerical example. Let V = 10 V and R_total = 100 Ω → I1 = 10 / 100 = 0.1 A. If resistance doubles to 200 Ω → I2 = 10 / 200 = 0.05 A, exactly half. The numerical check confirms the proportional reasoning. Why Other Options Are Wrong: Be the same: would require resistance unchanged or voltage doubled to compensate; not our case.Be doubled: happens if resistance halves at constant voltage, not when it doubles.Reduce source voltage: the source voltage is given as constant; the current change does not “reduce” the source. Common Pitfalls: Confusing proportional and inverse proportional relationships. Remember: at constant V, current varies inversely with resistance (I ∝ 1/R). At constant R, current varies directly with voltage (I ∝ V). Final Answer: be halved

Question 5

Series circuit open-circuit measurement: A 20 V DC source supplies an 8 Ω resistor in series with a lamp. If the lamp is removed (leaving an open circuit at the lamp socket), what voltage will a voltmeter read across the lamp socket terminals?

Accepted Answer

Introduction / Context: Measuring voltages in series circuits requires understanding where drops occur. When a component is removed, the circuit may become open, changing how voltage is distributed. This question reinforces the concept that with zero current, resistive drops vanish and the source appears across the open terminals. Given Data / Assumptions: DC source voltage: 20 V. Series elements: 8 Ω resistor and a lamp originally in series. The lamp is removed, creating an open circuit at the lamp socket. Voltmeter has ideally infinite input resistance and is connected across the lamp socket. Concept / Approach: Ohm’s law dictates the voltage drop across a resistor is V_R = I * R. In an open circuit, the loop current I is zero. Therefore, the drop across the 8 Ω resistor becomes 0 V, and the entire source voltage appears across the open terminals (the lamp socket). Step-by-Step Solution: Lamp removed → circuit open → I = 0 A.Resistor drop: V_R = I * R = 0 * 8 = 0 V.KVL around loop: source 20 V must appear across the only remaining break → lamp socket sees 20 V.Voltmeter reading across lamp socket = 20 V. Verification / Alternative check: Model the lamp socket as two open terminals. With no closed path, current is zero. Any series resistive element has zero drop, leaving the full source across the open—consistent with standard open-circuit behavior. Why Other Options Are Wrong: 0 V: would require a short across the socket, not an open.8 V or 12 V: arbitrary partial drops contradict zero-current condition (no drop across finite resistance with I = 0). Common Pitfalls: Forgetting that voltmeters do not provide a current path. A high-impedance meter does not “close” the circuit; thus no current flows and no series drop occurs. Final Answer: 20 V

Question 6

Flashlight battery polarity: A flashlight uses three 1.5 V cells in series. If one cell is inserted backward (reverse polarity), what happens to the flashlight output?

Accepted Answer

Introduction / Context: Series battery stacks add voltages algebraically, taking polarity into account. Many small flashlights are designed for three 1.5 V cells (nominal 4.5 V) to drive a bulb or LED plus driver. Reversing one cell changes the net series sum and can prevent operation entirely. Given Data / Assumptions: Three cells rated 1.5 V each, intended to be in series and oriented correctly. One cell is reversed, contributing negative polarity relative to the others. Flashlight electronics (bulb/LED driver) expect approximately 4.5 V. Idealized cells assumed for the polarity-sum explanation. Concept / Approach: Series voltages add with sign. With correct orientation, V_total = 1.5 + 1.5 + 1.5 = 4.5 V. With one reversed, V_total = 1.5 + 1.5 − 1.5 = 1.5 V. Most lamps or driver circuits designed for ~4.5 V will not operate properly at ~1.5 V; an incandescent filament may not glow perceptibly, and many LED drivers have undervoltage lockout or cannot maintain current regulation. Step-by-Step Solution: Compute nominal: three cells correct → 4.5 V.Reverse one cell: algebraic sum → 1.5 V total.Compare to required operating voltage: 1.5 V is far below design point.Conclusion: no visible output; flashlight is effectively off. Verification / Alternative check: Practical test: a 4.5 V incandescent bulb at 1.5 V draws much less current and produces negligible light; an LED driver typically shuts down under low input voltage. Both cases appear “off” to a user. Why Other Options Are Wrong: Brighter than normal / the same: contradict the reduced supply voltage.Dimmer than normal: while technically some small glow could occur with simple bulbs, in typical products the result is functionally off; the robust answer is off. Common Pitfalls: Ignoring polarity in series stacks. One reversed cell subtracts from the total, and in some electronics can also stress protection circuitry due to reverse biasing. Final Answer: off

Question 7

Voltage division in series: A 45 V DC source is applied to a series network with total resistance 3300 Ω. If one of the resistors, R3, is 1200 Ω, what is the voltage drop across R3?

Accepted Answer

Introduction / Context: Voltage division in series circuits is a fundamental technique for predicting how a source voltage distributes across individual elements. This problem directly applies the voltage divider rule to determine the drop across one known resistor when total resistance and source voltage are given. Given Data / Assumptions: Total series resistance, R_total = 3300 Ω. Source voltage, V_s = 45 V DC. Target resistor, R3 = 1200 Ω (in series). Ideal resistors; no additional elements. Concept / Approach: For series circuits, current I is the same through all resistors: I = V_s / R_total. The voltage across any resistor R_x is V_x = I * R_x. Combining, V_x = V_s * (R_x / R_total). This is the standard voltage divider formula. Step-by-Step Solution: Compute the ratio: R3 / R_total = 1200 / 3300 = 0.363636…Apply divider: V_R3 = 45 * 0.363636…Multiply: 45 * 0.363636… = 16.3636… V.Round sensibly to two decimals: V_R3 ≈ 16.36 V. Verification / Alternative check: Current method: I = 45 / 3300 = 0.013636… A. Then V_R3 = I * 1200 = 0.013636… * 1200 = 16.3636… V. Both methods agree. Why Other Options Are Wrong: 32.72 V: would correspond to a much larger fraction (about 0.727) not matching 1200/3300.10.90 V: too small; implies R3 is roughly 800 Ω of the total, which it is not.15.00 V: a rounded guess, not the precise divider result. Common Pitfalls: Forgetting that only series elements share the same current; mixing up series and parallel results leads to incorrect voltage predictions. Ensure total resistance includes all series elements when using the divider rule. Final Answer: 16.36 V

Question 8

These components are in series.

Accepted Answer

True

Question 9

Ground is the reference point from which voltages are measured in an electronic circuit.

Accepted Answer

True

Question 10

If 7.3 kΩ, 1.8 kΩ, and 4.9 kΩ resistors are wired in series, the total resistance is 14 kΩ.

Accepted Answer

True

Question 11

The total resistance of a series circuit is equal to the average of all the resistance values.

Accepted Answer

False

Question 12

To find the total resistance in series, add the individual resistances.

Accepted Answer

True

Question 13

The sum of the voltage drops across series resistors must equal the total voltage applied.

Accepted Answer

True

Question 14

If four 9 V batteries are connected series-aiding, the total voltage is 36 V.

Accepted Answer

True

Question 15

A short in a series circuit causes the total circuit current to decrease.

Accepted Answer

False

Question 16

An open in a series circuit will cause maximum voltage across the power supply terminals.

Accepted Answer

True

Question 17

Voltage sources are added when they are series-opposing.

Accepted Answer

False

Question 18

The total resistance of a series circuit always depends on the highest value resistor in that circuit.

Accepted Answer

False

Question 19

The total resistance is 35 Ω.

Accepted Answer

True

Question 20

The total power dissipated in a series resistive circuit exceeds the sum of the resistor powers.

Accepted Answer

False

Series Circuits Questions