Electric Charge, Current and Voltage Lesson Test

Answer: B

Correct answer: Ampere. Current is a rate: how much charge flows past a point each second, and one ampere is one coulomb per second. The usual mix-up is with the coulomb, which counts an amount of charge rather than how fast it moves. A water analogy helps: coulombs are like gallons, amperes are like gallons per minute, and volts are like the pressure pushing the flow.

• A: A coulomb measures charge quantity.
• B (correct): An ampere measures charge transferred per second.
• C: A volt measures potential difference.
• D: An ohm measures resistance.

Answer: C

Correct answer: Positive. An atom's nucleus holds protons, which are positive, and neutrons, which carry no charge, while electrons around the nucleus are negative. A proton and an electron carry equal-sized charges of opposite sign, which is why a balanced atom is neutral. Do not confuse charge sign with current type: alternating describes how a current changes direction over time, not a property of a particle.

• A: Alternating describes some currents, not charge sign.
• B: An electron, not a proton, has negative charge.
• C (correct): A proton has positive elementary charge.
• D: A neutron is neutral.

Answer: C

Correct answer: They repel each other. The basic rule of electrostatics is that like charges push apart and unlike charges pull together, whether both are positive or both are negative. The force grows stronger as the objects get closer. Charges do not cancel or disappear when objects interact, because charge is conserved. A practical example is two rubbed balloons that hang apart from each other because both picked up the same kind of charge.

• A: Electric repulsion does not imply merging.
• B: Charge is conserved, not automatically destroyed.
• C (correct): Two like net charges repel.
• D: Like charges do not attract in the simple electrostatic model.

Answer: D

Correct answer: Electric potential difference. One volt means one joule of energy for each coulomb of charge moved between two points, so voltage describes the electrical push behind current, not the current itself. Current, the charge passing per second, is measured in amperes, and resistance is measured in ohms. Remember that voltage is always measured between two points, which is why a voltmeter has two leads placed across a component.

• A: Resistance in a conductor is measured in ohms, not volts.
• B: Charge passing per second is current, measured in amperes.
• C: Current in a circuit is measured in amperes, not volts.
• D (correct): Voltage represents energy difference per unit charge.

Answer: D

Correct answer: Mobile electrons. In a metal, each atom releases one or more outer electrons into a shared sea that can drift through the fixed lattice of positive ions. When a voltage is applied, these free electrons drift slowly and carry the current. Protons are locked in nuclei and stay in place, and neutrons have no charge at all, so neither can carry current in a wire.

• A: Neutrons have no electrical charge.
• B: Nuclei ordinarily remain in the metal lattice.
• C: Protons do not leave the conductor as current carriers.
• D (correct): Mobile electrons are the main current carriers in metals.

Answer: B

Correct answer: 1 C each second. The ampere is defined as a rate of charge flow, so 1 A means one coulomb of charge passes a point every second. The tempting wrong choice is joules per second, but that describes power, measured in watts. Keeping the base units straight helps: charge in coulombs, current in coulombs per second, voltage in joules per coulomb, and power in joules per second.

• A: $\frac{\mathrm{J}}{\mathrm{s}}$ is a watt, a unit of power.
• B (correct): By definition 1 A = 1 $\frac{C}{s}$.
• C: $\frac{\mathrm{V}}{\mathrm{s}}$ describes a rate of voltage change, not current.
• D: Ω/s is a rate of resistance change.

Answer: A

Correct answer: Provide a complete conductive path. Current flows only around a closed loop, so a lamp lights only when there is an unbroken path from one battery terminal, through the lamp, and back to the other terminal. A switch is just a controllable gap in that loop. Opening it does not remove the battery's voltage; it only stops the flow. This is why a battery still reads its voltage with nothing connected.

• A (correct): Closing the switch completes the path for sustained current.
• B: Battery voltage can remain present.
• C: No electrons are destroyed by the switch.
• D: A switch does not change the battery into an insulator.

Answer: C

Correct answer: Coulomb. The coulomb counts a quantity of charge, roughly the charge of about 6.24 × 10¹⁸ electrons. It is easy to confuse with the ampere, but an ampere is a rate, one coulomb per second, just as a gallon is an amount and gallons per minute is a flow. Volts describe energy per coulomb, and watts describe energy per second, so neither measures charge itself.

• A: Watts measure power.
• B: Volts measure potential difference.
• C (correct): Coulombs measure charge. Additional check: A coulomb is an amount of electric charge, while an ampere is the rate of transporting that amount.
• D: Amperes measure charge rate.

Answer: A

Correct answer: negative. Ordinary objects become charged by gaining or losing electrons, since protons stay bound in atomic nuclei. Each extra electron adds a small negative charge, so the object becomes more negative; losing electrons leaves it more positive. A common error is to think charge is created by rubbing, but electrons are only moved from one material to another. Magnetism is a separate effect and does not describe charge sign.

• A (correct): Electrons are negative, so gaining them shifts charge negative.
• B: Magnetism is not the sign of electrostatic charge.
• C: Loss of electrons makes charge more positive.
• D: Gaining extra electrons does not guarantee neutrality.

Answer: C

Correct answer: Voltmeter. Because voltage is a difference between two points, a voltmeter is connected across, or in parallel with, the component being tested. An ideal voltmeter has very high resistance so it draws almost no current and does not disturb the circuit. The common mix-up is the ammeter, which must be placed in series so the current passes through it. An ohmmeter measures resistance, usually with the circuit powered off.

• A: An ammeter measures current through a path.
• B: A thermometer measures temperature.
• C (correct): A voltmeter reads voltage between its two leads.
• D: An ohmmeter measures resistance.

Answer: C

Correct answer: 3 A. Current is the amount of charge divided by the time it takes to pass,

Here 12 C spread over 4 s means 3 C pass each second, which is 3 A. Checking units helps you avoid mistakes: coulombs divided by seconds gives amperes, while multiplying them gives coulomb-seconds, which is not current. Dividing time by charge flips the ratio and gives a value that is too small.

• A: 48 is the product, not the ratio.
• B: Adding charge and time has no physical meaning.
• C (correct): I = $\frac{Q}{t}$ = $\frac{12}{4}$ = 3 A.
• D: 0.33 reverses the fraction.

Answer: D

Correct answer: 14 C. Since current is charge per second, total charge equals current multiplied by time,

At 2 A, two coulombs pass each second, so after 7 s a total of 14 C has passed. A quick unit check confirms it: amperes are coulombs per second, and multiplying by seconds leaves coulombs. Adding or dividing the numbers does not produce a unit of charge.

• A: 3.5 comes from dividing time by current.
• B: 9 comes from adding 2 and 7.
• C: 28 would require 4 A for 7 s.
• D (correct): Q = It = 2 × 7 = 14 C. Additional check: Since 1 A means 1 C each second, the calculation reports cumulative charge, not potential difference.

Answer: D

Correct answer: 30 J. Voltage tells you how much energy each coulomb gains or loses, so the total energy is voltage times charge,

Each of the 5 C carries 6 J, giving 30 J in all. Units confirm it: volts are joules per coulomb, and multiplying by coulombs leaves joules. Dividing or adding the values does not produce an energy unit, which is a quick way to rule those answers out.

• A: Dividing 6 by 5 is not the energy relation.
• B: Adding quantities with different units is invalid.
• C: The reciprocal has no requested energy unit.
• D (correct): E = VQ = 6 × 5 = 30 J.

Answer: B

Correct answer: 3 V. Voltage is defined as energy transferred per unit charge,

so you divide the joules by the coulombs. Here 24 J shared among 8 C gives 3 J for each coulomb, which is 3 V. Multiplying the values gives a unit of joule-coulombs that has no meaning here, and dividing the other way gives coulombs per joule. Writing the units first is the surest way to pick the correct operation.

• A: Multiplication produces energy times charge.
• B (correct): V = $\frac{E}{Q}$ = $\frac{24}{8}$ = 3 V.
• C: $\frac{8}{24}$ inverts the ratio.
• D: Subtracting 8 from 24 is not the definition.

Answer: A

Correct answer: From positive toward negative through the load. Conventional current was defined before electrons were discovered, as the direction positive charge would move. Outside the source it flows from the positive terminal, through the load, to the negative terminal. In metal wires, the actual carriers are electrons, which drift the opposite way, from negative to positive. Both descriptions predict the same circuit behavior; diagrams and diode arrows use the conventional direction.

• A (correct): The conventional current arrow points from + to − outside the source.
• B: That is the direction of electron drift in a metal wire.
• C: Circuit orientation on the page does not set current direction.
• D: Conventional current forms a complete circuit, not a midpoint-only path.

Answer: B

Correct answer: 0.5 A. The prefix milli means one thousandth, so 1 mA = 0.001 A, and converting to amperes means dividing by 1,000. That turns 500 mA into 0.5 A. A quick sense check: a smaller unit needs a larger number to express the same amount, so the value in amperes must be smaller than 500. Shifting the decimal point three places to the left handles any milli-to-base conversion.

• A: 5 A is ten times too large.
• B (correct): 500 mA = $\frac{500}{1000}$ A = 0.5 A.
• C: The prefix milli reduces rather than increases the value.
• D: 0.05 A corresponds to 50 mA.

Answer: B

Correct answer: Steady loop current stops even if a potential difference remains. A battery maintains a voltage between its terminals through chemical reactions, whether or not anything is connected. Current, however, needs a complete loop, so an open switch breaks the path and the flow stops. The full battery voltage then appears across the open switch contacts. This is why a voltmeter can read a battery's voltage even when the lamp in its circuit is off.

• A: An open switch does not measure current.
• B (correct): Breaking the path interrupts steady current, though voltage can remain.
• C: A battery can maintain potential difference with an open circuit.
• D: Opening a switch does not eliminate charged particles.

Answer: A

Correct answer: Charge redistribution in the paper can produce attraction. The balloon's charge pulls opposite charges within the paper's molecules slightly closer and pushes like charges slightly away, a process called polarization. Since the nearer, opposite charges feel a stronger force than the farther, like charges, the net effect is attraction even though the paper stays neutral overall. That is why a neutral object can be attracted, but two objects can only repel if both are charged.

• A (correct): Polarization separates charge slightly within neutral matter.
• B: Electric attraction does not create protons.
• C: Gravity does not change sign because of static charging.
• D: Neutral objects need not carry opposite net charge.

Answer: B

Correct answer: In series with the lamp. To measure current, the current must actually flow through the meter, so you break the circuit and insert the ammeter in the same path as the lamp. An ideal ammeter has almost zero resistance, so it does not change the current. That same low resistance is why connecting an ammeter across a battery or a component creates a near short circuit, which can damage the meter or the source.

• A: Ammeter across the source can produce a short circuit.
• B (correct): Series connection carries the same branch current through the meter.
• C: Disconnected instruments cannot read branch current.
• D: A meter across the lamp bypasses it instead of carrying the lamp current.

Answer: A

Correct answer: 2 A. Average current is the total charge moved divided by the time taken,

Four coulombs delivered over 2 s means 2 coulombs per second on average, which is 2 A. Writing units makes the operation clear: coulombs over seconds gives amperes. Multiplying charge and time or dividing time by charge produces units that are not current, so those results can be ruled out.

• A (correct): I = $\frac{Q}{t}$ = $\frac{4}{2}$ = 2 A. Additional check: A current measured in A must equal transferred coulombs divided by seconds, not their product.
• B: Adding values lacks a unit basis.
• C: Multiplication gives the wrong operation.
• D: $\frac{t}{Q}$ inverts the relation.

Answer: C

Correct answer: About 9 J of energy per coulomb of charge between its terminals. Voltage is energy per unit charge, so a 9 V rating means each coulomb moved through an external circuit gains about 9 J. The rating does not set current or power; those depend on the load. A small resistance draws more current than a large one from the same battery. Storage capacity is a separate rating, usually given in milliampere-hours.

• A: Power depends on both voltage and current, so it changes with the load.
• B: Charge capacity is a separate rating and is not given by the voltage.
• C (correct): V = $\frac{J}{C}$; the nominal terminal voltage expresses energy per charge.
• D: Current depends on the load and circuit conditions, not just the voltage rating.

Answer: A

Correct answer: Mobile charges generally move more readily through metal. Metals have free electrons that drift easily, while rubber binds its electrons tightly to its molecules, so very few can move. The difference in conductivity is enormous, which is why rubber and plastic coat wires and tools. A common misconception is that conductors cannot hold static charge; an insulated metal object can, but the charge spreads quickly over its surface.

• A (correct): Mobile charges travel more readily in most metals.
• B: Metal objects can hold net charge under some conditions.
• C: Their conductivities differ enormously under normal conditions.
• D: Rubber is not an electrical source merely because it is an insulator.

Answer: D

Correct answer: 3 A. On a charge-versus-time graph, the slope is rise over run, change in charge divided by change in time, which is exactly the definition of current. Here 18 C over 6 s gives 3 C/s, which is 3 A. The units of a slope are always the y-axis unit divided by the x-axis unit, which gives an easy check. Flipping the ratio gives seconds per coulomb, not current.

• A: Resistance cannot be derived from this graph alone.
• B: Multiplying does not provide the graph slope, and volts are incorrect.
• C: $\frac{6}{18}$ reverses the rise/run.
• D (correct): 18 C / 6 s = 3 $\frac{C}{s}$ = 3 A.

Answer: A

Correct answer: 15 W. Power is the rate of energy transfer, and for an electrical load it equals voltage times current,

Each coulomb carries 5 J, and 3 coulombs arrive each second, so 15 J are delivered per second, which is 15 W. Dividing voltage by current gives resistance in ohms, not power, a common error when formulas are mixed up.

• A (correct): P = VI = 5 × 3 = 15 W. Additional check: Volts multiplied by amperes gives joules per second, which is called a watt.
• B: This divides voltage by current.
• C: This divides amperes by volts.
• D: This adds quantities with unlike units.

Answer: A

Correct answer: 5 V. Potential difference is energy transferred per unit charge, so

Dividing 15 J by 3 C gives 5 J/C, and one joule per coulomb is one volt. The unit labels on the other choices are a clue: watts, amperes, and coulombs are not units of potential difference. A good habit is to check that the unit of your answer matches the quantity asked for.

• A (correct): V = 15 J/3 C = 5 $\frac{J}{C}$ = 5 V.
• B: The reciprocal is not potential difference.
• C: 45 is a product with an incorrect unit.
• D: Subtraction yields no valid charge answer.

Answer: B

Correct answer: 0.8 A. Charge is not used up in a circuit, so in a single loop the same current flows through every point, before and after each component. What the lamp uses is energy, which shows up as a voltage drop across it, not a loss of current. The idea that loads consume electrons is a common misconception; electrons return to the source and are pushed around again.

• A: The lamp does not halve the conserved charge flow by itself.
• B (correct): The same steady current passes every point of an unbranched path.
• C: An energized, unbranched loop carries a nonzero current.
• D: Loads transfer energy; they do not consume electrons.

Answer: B

Correct answer: 0.00025 A. The prefix micro, written µ, means one millionth, so 1 µA = 0.000001 A. Converting 250 µA to amperes means dividing by 1,000,000, which moves the decimal point six places to the left. A common slip is to treat micro like milli, which divides by only 1,000 and gives a value a thousand times too large. Microamp currents are typical in sensitive sensors and electronic circuits.

• A: This treats micro as milli.
• B (correct): $\frac{250}{1}$,000,000 = 0.00025 A.
• C: This is one hundred times too large.
• D: This reverses the metric-prefix conversion.

Answer: C

Correct answer: AC changes direction periodically, while DC does not reverse direction. The defining feature is direction: alternating current reverses many times per second, 60 cycles per second on U.S. household power, while direct current always flows one way. A DC current can still vary in strength, as in a pulsing or rectified supply, and remain DC as long as it never reverses. Batteries supply DC, while wall outlets supply AC.

• A: Batteries are familiar DC sources.
• B: AC reverses direction, and a DC current can vary in strength as long as it does not reverse.
• C (correct): Direction reversal is the crucial distinction.
• D: Both AC and DC involve moving charge, and both deliver energy to loads.

Answer: A

Correct answer:

Start from the definition of current,

and solve for time by multiplying both sides by t and dividing by I. A unit check confirms the result: coulombs divided by coulombs per second leaves seconds. The inverted form I/Q would give units of one per second, and the product gives coulomb-squared per second, so neither can be a time.

• A (correct): From I = $\frac{Q}{t}$, rearrange to t = $\frac{Q}{I}$.
• B: $\frac{V}{Q}$ does not express elapsed time.
• C: The product gives incorrect dimensions.
• D: The inverse ratio is not time.

Answer: D

Correct answer: Charge is redistributed rather than created. Conservation of charge says the total charge of an isolated system never changes. When two materials are rubbed together, electrons move from one surface to the other, so one becomes negative and the other becomes equally positive. The amounts are always equal and opposite, so the total stays the same. Protons stay bound in atomic nuclei, so they are not what moves in ordinary friction charging.

• A: Rubbing does not double the system’s charge.
• B: The mobile transferred particles are ordinarily electrons, not protons.
• C: One can gain electrons while the other loses electrons.
• D (correct): Total electric charge stays constant; electrons are transferred.

Answer: D

Correct answer: 18 C. Charge equals current times time, but the units must match first: amperes are coulombs per second, so time must be in seconds and current in amperes. Convert 150 mA to 0.150 A and 2 minutes to 120 s, then multiply to get 18 C. The most common errors are leaving time in minutes or current in milliamperes, which gives answers off by factors of 60 or 1,000.

• A: 0.150 A × 2 s ignores conversion of minutes to seconds.
• B: This treats 150 mA as 150 A or mishandles time.
• C: This is an unrelated numerical ratio.
• D (correct): Q = 0.150 A × 120 s = 18 C.

Answer: A

Correct answer: 6 V and 4 s. Treat this as two separate definitions. Voltage is energy per charge, so 72 J divided by 12 C gives 6 V. Current is charge per time, so time is charge divided by current: 12 C divided by 3 A gives 4 s. Check that each number carries the correct unit; swapping volts and amperes or seconds is a frequent error when several quantities appear in one problem.

• A (correct): V = $\frac{72}{12}$ = 6 V; t = $\frac{12}{3}$ = 4 s.
• B: The numerical values are paired with wrong quantities and units.
• C: 72−12 and 12×3 are invalid substitutions.
• D: This confuses energy/current and time.

Answer: D

Correct answer: 3.2 A. Average current over a whole interval is total charge divided by total time, not the average of the separate rates. Add the charge, 10 C plus 6 C gives 16 C, and the time, 2 s plus 3 s gives 5 s, then divide to get 3.2 A. Averaging 5 A and 2 A gives 3.5 A, which is wrong because the two intervals last different lengths of time.

• A: 5 A is only the first-interval rate ($\frac{10}{2}$).
• B: 5.3 A comes from dividing the total 16 C by only the last 3 s interval.
• C: 2 A is only the second-interval rate ($\frac{6}{3}$).
• D (correct): Total charge = 16 C over 5 s, so average I = 3.2 A.

Answer: C

Correct answer: 4320 C. An ampere-hour is the charge delivered by 1 A flowing for one hour. Since 1 A is 1 C per second and an hour has 3,600 s, 1 Ah equals 3,600 C. Multiply 1.2 by 3,600 to get 4,320 C. Battery capacity ratings in mAh work the same way: 1 mAh equals 3.6 C. The rating describes charge, not voltage or energy.

• A: Ah is not numerically equal to C.
• B: This is too small by a factor of six.
• C (correct): 1 Ah = 3600 C, so 1.2 × 3600 = 4320 C.
• D: This does not correctly convert hours to seconds.

Answer: C

Correct answer: 20 J. First find power, the energy delivered each second, from

Convert 400 mA to 0.400 amperes, so the power is 5 V × 0.400 A = 2 W. Then multiply by time, because energy equals power times duration: 2 W × 10 s = 20 J. Forgetting the time step leaves a power value, and forgetting the milliamp conversion makes the answer a thousand times too large.

• A: This ignores multiplication by the 10 s duration.
• B: This treats 400 mA as 400 A and also omits the 10 s duration.
• C (correct): P = 5 V × 0.400 A = 2 W; E = Pt = 20 J.
• D: 10 s × 5 V cannot give energy without current.

Answer: B

Correct answer: Twice as much. Ohm's law,

shows that with resistance held constant, current is directly proportional to voltage. Doubling the voltage from 4 V to 8 V therefore doubles the current. Four times would apply to power, since P = V²/R, which is a common mix-up. Direct proportion means any change in voltage produces the same factor of change in current for a fixed resistor.

• A: $\frac{8}{4}$ = 2, not 4.
• B (correct): For fixed R, I = $\frac{V}{R}$; doubling V doubles I.
• C: The currents differ because voltages differ.
• D: Higher V across equal fixed R means higher, not lower, I.

Answer: C

Correct answer: The sphere’s mobile electrons redistribute, but its total charge remains zero. The negative rod repels free electrons in the metal, pushing them to the far side and leaving the near side positive; this is charging by induction. Because the sphere is isolated and nothing touches it, no charge enters or leaves, so its net charge stays zero. A lasting net charge appears only if the sphere is grounded while the rod is nearby.

• A: Without contact or a conducting path, no charge transfer is required.
• B: Positive nuclei do not migrate throughout the metal in this setting.
• C (correct): Induction redistributes mobile charge without changing isolated sphere net charge.
• D: Conservation of charge forbids creation here.

Answer: D

Correct answer: A potential difference may still exist across the open switch. Zero current only means the loop is broken, not that voltage is gone. With an ideal open switch, the battery's full voltage appears across the switch gap, since no current flows to cause voltage drops elsewhere. This matters for safety: a dead circuit must be verified with a voltmeter, not assumed from the absence of current, because opened circuits can still carry dangerous voltage.

• A: Electrons remain in material even when no net current flows.
• B: Energized parts can remain hazardous despite no loop current.
• C: Open circuit does not imply the battery loses its emf.
• D (correct): An open gap can sustain a potential difference without steady current.

Answer: D

Correct answer: 4 A and 4 V. Apply the two definitions separately. Current is charge per time, so 36 C over 9 s gives 4 A. Voltage is energy per charge, so 24 J for 6 C gives 4 J/C, or 4 V. The matching numbers are a coincidence, so do not let them confuse you. Writing each ratio with units, C/s and J/C, keeps you from pairing a value with the wrong quantity.

• A: $\frac{36}{9}$ is 4 A, not 9 A.
• B: Neither ratio equals 6 in this setup.
• C: Energy per charge is $\frac{24}{6}$ = 4 V, not 24 V.
• D (correct): I = $\frac{36}{9}$ = 4 A; V = $\frac{24}{6}$ = 4 V.

Answer: B

Correct answer: It halves from 0.5 A to 0.25 A. With the voltage fixed, Ohm's law,

shows that current is inversely proportional to resistance. At 10 Ω the current is

doubling the resistance to 20 Ω cuts it to 0.25 A. A larger resistance reduces current but does not stop it, since only an open circuit, with effectively infinite resistance, drops the current to zero.

• A: Ohm’s law links I to resistance as well as voltage.
• B (correct): $\frac{5}{10}$ = 0.5 A; $\frac{5}{20}$ = 0.25 A.
• C: At fixed voltage, increasing resistance decreases I.
• D: Doubling finite resistance does not necessarily make I zero.

Answer: D

Correct answer: Net positive, because electrons were pushed off to ground and could not return. The negative rod repels the sphere's mobile electrons toward the far side. Grounding that side lets some of those electrons escape to Earth. Removing the ground before the rod traps the shortage, so the sphere ends with more protons than electrons: a net positive charge, opposite to the rod.

• A: The rod never touched the sphere, so none of its charge transferred; this describes charging by contact, not induction.
• B: Without the ground connection it would stay neutral, but grounding let electrons leave, so the total changed.
• C: A negative rod repels electrons, so they flow away into the ground, not up from it.
• D (correct): The rod drove electrons out through the ground wire; once the ground is removed first, the sphere is left short of electrons.

Answer: A

Correct answer: From the negative terminal, through the bulb, toward the positive terminal. Conventional current is defined as leaving the positive terminal, but the carriers in a metal wire are electrons, which are negative. They drift the opposite way: out of the negative terminal, through the bulb, and back toward the positive terminal.

• A (correct): Electrons are negative, so they drift opposite to conventional current, from negative toward positive through the load.
• B: This is the direction of conventional current, which is the opposite of the real electron drift in a metal.
• C: That describes household AC; a battery supplies DC, which does not reverse.
• D: Charge flows around one loop; nothing is used up inside the bulb, so carriers cannot simply meet and stop there.

Answer: D

Correct answer: 35 mA. One ampere equals 1,000 mA, so to change amperes to milliamperes multiply by 1,000: 0.035 times 1,000 is 35 mA. A smaller unit always needs a larger number.

• A: This slips one decimal place; 0.035 A times 1,000 is 35, not 3.5.
• B: This moves the decimal four places instead of three; 350 mA would be 0.35 A.
• C: This divides by 1,000 instead of multiplying; going from amperes to the smaller milliampere unit must give a bigger number.
• D (correct): Multiplying 0.035 A by 1,000 gives 35 mA.

Answer: A

Correct answer: 26,640 J. First turn the capacity into charge: 1 Ah is 3,600 C, so 2 Ah is 7,200 C. Then use

• A (correct): 2 Ah is 7,200 C, and each coulomb receives 3.7 J, so 7,200 times 3.7 gives 26,640 J.
• B: 7,200 is the charge in coulombs; it still has to be multiplied by the voltage to give energy.
• C: 2 Ah times 3.7 V is 7.4 watt-hours, not joules; the hours must be turned into seconds.
• D: This uses 60 s per hour instead of 3,600 s, making the charge far too small.

Answer: A

Correct answer: +3 nC each. Charge is conserved, so first find the total, keeping the signs: +8 nC + (−2) nC = +6 nC. Touching only moves electrons between the spheres, and identical spheres split the total evenly: 6 divided by 2 is +3 nC each. Check:

nC, the same total as before.

• A (correct): The total is +8 + (−2) = +6 nC, and splitting it equally gives +3 nC on each sphere.
• B: This adds the sizes 8 and 2 while ignoring the minus sign, which would create charge that was never there.
• C: This just halves each starting charge, so the spheres would still differ; touching makes them share one total.
• D: The charges do not cancel completely because they are unequal; conservation leaves a net +6 nC to share.

Answer: B

Correct answer: The lamp goes dark and the meter reads nearly the full battery voltage. A voltmeter is built with very high resistance so it can sit across a part without drawing current. Placed in series, that high resistance becomes part of the only path, so the current drops almost to zero and the lamp goes out. With almost no current through the lamp, the lamp has almost no voltage drop, so the meter shows nearly the whole battery voltage.

• A: Almost no current flows, but nearly the whole battery voltage then appears across the meter itself; zero current does not mean zero voltage.
• B (correct): A voltmeter has very high resistance, so in series it blocks almost all the current; the lamp goes dark and nearly all the battery voltage appears across the meter.
• C: The meter would only read the lamp's voltage if its leads were placed across the lamp, in parallel.
• D: That is the danger of an ammeter placed across a source; a voltmeter's high resistance does the opposite.

Answer: D

Correct answer: 1.25 × 10¹⁶. In one second the charge passing is

Each electron carries $1.6\times 10^{-19}$ C, so divide.

• A: This treats 2 mA as 2 A; the milli prefix makes the charge 1,000 times smaller.
• B: This multiplies the charge by e instead of dividing by it, giving a meaninglessly tiny number.
• C: This is the electron count for 0.5 C; it divides by 2 instead of using the 0.002 C that actually passes.
• D (correct): In 1 s, 0.002 C passes, and 0.002 divided by 1.6 times 10 to the negative 19 gives about 1.25 times 10 to the 16.

Answer: B

Correct answer: 3 A. Power, voltage and current are linked by

Check: 12 V times 3 A gives back 36 W.

• A: This multiplies the power by the voltage; current comes from dividing power by voltage.
• B (correct): From P = VI, the current is 36 W divided by 12 V, which is 3 A.
• C: This divides voltage by power, which is the formula turned upside down.
• D: Adding 36 and 12 combines unlike quantities and has no meaning.

Answer: C

Correct answer: 150 s. Convert first: 40 mA is 0.04 A. Then rearrange

to

That is 2.5 minutes, a sensible time for a small current to move several coulombs.

• A: This multiplies charge by current; time is charge divided by current.
• B: This divides 6 by 40 without converting 40 mA to 0.04 A, so it is off by a factor of 1,000.
• C (correct): Converting 40 mA to 0.04 A and using t = Q/I gives 6 divided by 0.04, which is 150 s.
• D: This multiplies 6 by 40 instead of dividing, ignoring the milli prefix as well.

Answer: C

Correct answer: The lamp goes dark while the current in the wires rises sharply. The wire is a short circuit: an unintended low-resistance path around the load. Almost all the current takes that easy path, so the lamp goes dark, and with so little resistance in the loop the current becomes very large, heating the wires and draining the battery. Fuses and breakers exist to open the circuit in this situation.

• A: The new wire gives current an easier path around the lamp, not into it, so the lamp receives little or none.
• B: A short adds a path rather than breaking one; current increases instead of stopping.
• C (correct): The low-resistance wire bypasses the lamp, so current skips the lamp and grows large, heating the wires.
• D: Current takes the path available to it, and a near-zero-resistance wire carries most of it.