Ohm's Law Lesson Test: 40 Electronics Questions

Answer: B

Correct answer: The rate at which charge flows past a point. The ampere is the unit of current. One ampere means one coulomb of charge passes a point every second, so it measures a rate of charge flow.

• A: That describes potential difference, which is measured in volts.
• B (correct): An ampere is one coulomb of charge passing a point each second, which is the definition of current.
• C: That describes resistance, which is measured in ohms.
• D: That describes power, which is measured in watts.

Answer: C

Correct answer:

Ohm's law says voltage equals current times resistance,

The other forms come from rearranging it:

• A: Ohm's law is a product, not a sum; volts cannot be found by adding amperes and ohms.
• B: Current divided by voltage is conductance; resistance is voltage divided by current.
• C (correct): Voltage equals current multiplied by resistance, the standard form of Ohm's law.
• D: Current is voltage divided by resistance, so multiplying them gives a meaningless result.

Answer: B

Correct answer: 0.45 A. Milli means one thousandth, so divide by 1,000 to change milliamperes to amperes: 450 mA ÷ 1,000 = 0.45 A.

• A: This divides by only 100, moving the decimal one place too few.
• B (correct): Dividing 450 mA by 1,000 gives 0.45 A.
• C: This divides by 10,000, moving the decimal one place too many.
• D: This multiplies by 1,000 instead of dividing, converting in the wrong direction.

Answer: C

Correct answer: 5 A. Use

Check: 5 A times 6 ohms gives back 30 V.

• A: This adds volts and ohms, which are different quantities and cannot be added.
• B: This multiplies voltage by resistance instead of dividing.
• C (correct): Dividing 30 V by 6 ohms gives 5 A.
• D: This divides resistance by voltage, which is upside down.

Answer: B

Correct answer: Ohm (Ω). Resistance is the opposition to current, and Ohm's law defines it as

so one ohm is one volt per ampere. Each quantity in V = IR has its own unit: volts for potential difference, amperes for current and ohms for resistance. The watt is easy to confuse here because it also appears in circuit formulas, but it measures power, which equals V × I. The Greek omega symbol marks resistance on diagrams and meters.

• A: A watt measures energy transfer per second, or power.
• B (correct): The ohm is the SI derived unit of electrical resistance.
• C: A volt measures electric potential difference.
• D: An ampere measures current rather than resistance.

Answer: C

Correct answer: 30 V. Voltage is current times resistance:

• A: This divides current by resistance, which gives neither voltage nor resistance.
• B: This divides resistance by current instead of multiplying.
• C (correct): Multiplying 2 A by 15 ohms gives 30 V.
• D: This adds amperes and ohms instead of multiplying them.

Answer: C

Correct answer: 1,000 milliamperes. The prefix milli means one thousandth, so a milliampere is 0.001 A and it takes 1,000 of them to make one ampere. To convert amperes to milliamperes, multiply by 1,000; to go back, divide by 1,000. Distractors that mix in volts or ohms change the kind of quantity, and a ratio per second describes a rate of change rather than an equal amount. The same milli pattern appears in millivolts and milliwatts.

• A: A millivolt refers to voltage, not current.
• B: Volt per ohm squared is not the definition of an ampere.
• C (correct): One ampere equals one thousand milliamperes.
• D: This expression is not a unit conversion for amperes.

Answer: D

Correct answer: 2,000 Ω. The metric prefix kilo means one thousand, so multiply the number in front of kΩ by 1,000. Going from a larger unit to a smaller one always makes the number bigger, which is a quick way to catch errors: an answer smaller than 2 cannot be right. Resistor markings commonly use k for thousands and M for millions of ohms, so 4.7 kΩ means 4,700 Ω and 1 MΩ means 1,000,000 Ω.

• A: This is smaller by a factor of one hundred.
• B: This mistakenly divides by one thousand.
• C: This is smaller by a factor of ten.
• D (correct): Kilo is one thousand: 2 kΩ equals 2,000 Ω.

Answer: B

Correct answer: The current is the same through both resistors. In series there is only one path, so charge cannot pile up or leak away between parts. Every series component carries the same current; what differs is the voltage drop, which is larger across the 100 ohm resistor.

• A: Larger resistance does not draw more current, and in series the current cannot differ anyway.
• B (correct): Series parts form a single path, so the same current passes through each one regardless of its resistance.
• C: Unequal currents happen in parallel branches; a single series path cannot have different currents.
• D: Current is not consumed; the current leaving a resistor equals the current entering it.

Answer: A

Correct answer: In series, so all of the lamp current passes through it. An ammeter measures current through a path, so the circuit is opened at one point and the meter is inserted in series. Because an ammeter has nearly zero resistance, it must never be placed across a source or a component.

• A (correct): Current is a flow through a path, so the circuit is opened and the meter inserted in that path.
• B: That describes an ohmmeter measurement, which cannot show the current the lamp draws in operation.
• C: An ammeter has almost no resistance, so placing it straight across a source acts like a short circuit.
• D: That is how a voltmeter is connected; a low-resistance ammeter there would bypass the lamp.

Answer: D

Correct answer: 6 mA. Ohm's law gives

but the resistance must be in plain ohms first: 1.5 kΩ is 1,500 Ω. Dividing 9 by 1,500 gives 0.006 A, and multiplying by 1,000 expresses it as milliamperes. A handy shortcut: volts divided by kilohms gives milliamperes directly, so

mA. Forgetting the kilo prefix inflates the answer a thousandfold, which is why 6 A is the most common wrong result.

• A: This value is one hundred times too large.
• B: The kilo-ohm prefix was ignored by a factor of one thousand.
• C: This current would correspond to 15 kΩ, not 1.5 kΩ.
• D (correct): 9 V / 1,500 Ω = 0.006 A = 6 mA.

Answer: B

Correct answer: 80 Ω. Ohm's law needs consistent units, so express the current in amperes before dividing: 300 mA is 0.3 A because milli means one thousandth. Then

A quick sanity check is to multiply back: 0.3 A × 80 Ω = 24 V. A common slip is multiplying voltage by current, which gives power in watts rather than resistance, and another is using 300 directly, which gives a resistance far too small.

• A: This value produces 30 mA rather than 300 mA.
• B (correct): Convert 300 mA to 0.3 A, then R = $\frac{24}{0.3}$ = 80 Ω.
• C: Multiplying 24 by 0.3 gives power in watts, not resistance.
• D: This would give 3 A at 24 V, ten times too much current.

Answer: C

Correct answer: 1.25 A. With resistance constant, current is proportional to voltage. The voltage ratio is

so the current is 0.5

Finding R first (

) and then 20 ÷ 16 gives the same answer.

• A: This multiplies the original current by the new voltage instead of by the voltage ratio.
• B: This multiplies by 8/20, scaling the current in the wrong direction.
• C (correct): The voltage increases by 20/8 = 2.5 times, so the current also increases 2.5 times: 0.5 × 2.5 = 1.25 A.
• D: With fixed resistance the current changes in proportion to voltage; it does not stay the same.

Answer: B

Correct answer: 12 A. At constant voltage, current is inversely proportional to resistance. The original load is

the new one is 1.5 ohms, so

twice the original.

• A: This halves the current, but lower resistance at fixed voltage makes current larger, not smaller.
• B (correct): Halving the resistance at the same voltage doubles the current: 6 A × 2 = 12 A.
• C: This quadruples the current, as if the resistance had been cut to one fourth.
• D: The current depends on resistance, so changing the load must change the current.

Answer: A

Correct answer: 1 A. Combine the factors separately: the current changes by

• A (correct): Voltage factor 2 divided by resistance factor 4 gives 0.5, so the current becomes 2 A × 0.5 = 1 A.
• B: This applies only the fourfold resistance and ignores the doubled voltage.
• C: This multiplies by both factors, but a larger resistance must reduce current.
• D: This applies only the doubled voltage and ignores the larger resistance.

Answer: A

Correct answer: 30 Ω. On a current-versus-voltage graph the slope is $1/R$, so take voltage over current at the point.

• A (correct): Resistance is voltage divided by current: 12 V ÷ 0.4 A = 30 ohms.
• B: This drops a decimal place in the current, dividing 12 by 4 instead of 0.4.
• C: This is the slope, 0.4/12, which on this graph is 1/R in siemens, not the resistance.
• D: This multiplies the voltage by the current instead of dividing.

Answer: D

Correct answer: 2 V: 0.05 A; 4 V: 0.10 A; 8 V: 0.20 A. Divide each voltage by its current. Only one set gives the same ratio every time:

A constant V/I ratio means a straight line through the origin, which defines an ohmic part.

• A: The V/I ratio climbs from 40 to 50 to about 67 ohms, the way a heating lamp filament behaves.
• B: The V/I ratio falls from 40 to about 27 to 20 ohms, so the resistance is not constant.
• C: The current stays fixed while voltage changes, so V/I goes from 20 to 80 ohms and is not constant.
• D (correct): Every reading gives V/I = 40 ohms, so the resistance is constant and the part is ohmic.

Answer: B

Correct answer: 12 V. Add the series resistances:

The 12 ohm resistor drops the remaining 4 V, and 12 V + 4 V = 16 V.

• A: This is the drop across the 12 ohm resistor, not the 36 ohm one.
• B (correct): Total resistance is 48 ohms, the current is 1/3 A, and 1/3 A × 36 ohms = 12 V.
• C: This uses the full supply voltage for one resistor that receives only part of it.
• D: This splits the voltage equally, which only happens when the series resistances are equal.

Answer: A

Correct answer: 150 Ω. With V on the vertical axis, the slope of the line is resistance. Convert 40 mA to 0.04 A first.

• A (correct): On a V-versus-I graph the slope is resistance: 6 V divided by 0.04 A is 150 ohms.
• B: This divides 6 by 40 without first converting 40 mA to 0.04 A, so it is off by a factor of 1,000.
• C: This divides current by voltage, which gives the reciprocal of resistance rather than resistance.
• D: This multiplies voltage by the current reading instead of dividing voltage by current.

Answer: B

Correct answer: In parallel, with one probe on each end of the resistor. A voltmeter measures the potential difference between two points, so it is connected in parallel, with its probes on either side of the component while the circuit is operating. Its high internal resistance lets it do this while drawing almost no current.

• A: Both probes on one point measure no difference, so the meter would read about zero.
• B (correct): Voltage is a difference between two points, so the meter goes across the resistor without breaking the circuit.
• C: A voltage drop appears only while current flows; with the power off the reading would be zero.
• D: Series insertion is how an ammeter is connected; a high-resistance voltmeter in series would block most of the current.

Answer: D

Correct answer: 0.8 A. Parallel branches share the full 12 V. The 20 ohm branch draws

A and the 60 ohm branch draws

The source supplies the sum, 0.8 A.

• A: This is only the current in the 60 ohm branch; branch currents must be added.
• B: This is only the current in the 20 ohm branch; branch currents must be added.
• C: This treats the resistors as if they were in series (80 ohms total).
• D (correct): Each branch gets 12 V: 12/20 = 0.6 A and 12/60 = 0.2 A, which add to 0.8 A.

Answer: A

Correct answer: 12 W. Use the form that fits voltage and resistance.

Check: I = 0.5 A, and 24 V × 0.5 A = 12 W.

• A (correct): Squaring the voltage and dividing by resistance gives 576 ÷ 48 = 12 W.
• B: This multiplies voltage by resistance, which is not a power formula.
• C: This divides resistance by voltage, which gives neither current nor power.
• D: 0.5 is the current in amperes; the right digits for current were given the wrong unit.

Answer: C

Correct answer: 200 Ω. Convert milliamperes to amperes by dividing by 1,000, then apply

An equivalent shortcut is that volts divided by milliamperes gives kilohms:

kΩ, which is 200 Ω. Check by multiplying back: 0.04 A × 200 Ω = 8 V. Small currents point to larger resistances, so an answer like 5 Ω or 20 Ω should look suspicious for a current of only a few hundredths of an ampere.

• A: At 8 V, 20 Ω would yield 400 mA, not 40 mA.
• B: This is the product of unconverted numerical values.
• C (correct): 40 mA = 0.04 A and $\frac{8}{0.04}$ = 200 Ω.
• D: This divides the unconverted 40 (mA) by 8 V, inverting the ratio.

Answer: D

Correct answer: 20 mA. Start by converting the prefixed resistance to ohms: 0.5 kΩ means 0.5 × 1,000, or 500 Ω. Then

Volts divided by kilohms also gives milliamperes directly:

mA. The wrong answers come from mishandling the decimal or prefix, each shifting the result by a power of ten. A useful habit is checking the product: 0.02 A × 500 Ω returns exactly the 10 V applied.

• A: A 0.2 A current implies 50 Ω, not 500 Ω.
• B: 200 mA equals 0.2 A and is ten times too high.
• C: A 2 A current requires just 5 Ω at 10 V.
• D (correct): 0.5 kΩ = 500 Ω; $\frac{10}{500}$ = 0.02 A = 20 mA.

Answer: B

Correct answer: 9 V. An open switch makes the current zero. With I = 0, each resistor's drop is

so none of the 9 V is used by the resistors and all of it appears across the open switch.

• A: Zero current does not mean zero voltage; the open gap holds the whole source voltage.
• B (correct): With no current, the resistors drop no voltage, so the full source voltage appears across the open gap.
• C: An open switch cannot produce more voltage than the source supplies.
• D: No current flows, so the resistors do not share the voltage with the gap.

Answer: D

Correct answer: 400 Ω. Convert first: 15 mA = 0.015 A. Then use

• A: This divides the current by the voltage, which is upside down.
• B: This divides 6 by 15 without converting milliamperes to amperes.
• C: This multiplies voltage by current instead of dividing.
• D (correct): Converting 15 mA to 0.015 A and dividing gives 6 ÷ 0.015 = 400 ohms.

Answer: A

Correct answer: 5 μA. Convert: 4 megohms = 4,000,000 ohms.

• A (correct): Volts divided by megohms gives microamperes: 20 ÷ 4 = 5 microamperes.
• B: Volts divided by kilohms gives milliamperes; megohms are a thousand times larger.
• C: This multiplies voltage by resistance instead of dividing.
• D: This ignores the mega prefix and divides 20 by 4 ohms.

Answer: C

Correct answer: 27 W. With R fixed,

so power scales with the square of current. The current ratio is

so power becomes

times larger: 3

Check:

• A: This doubles the power, which matches neither the threefold current change nor the square law.
• B: This triples the power along with the current, but power depends on the square of current.
• C (correct): The current triples, so power rises by 3 squared, or 9 times: 3 W × 9 = 27 W.
• D: This squares the 9 W figure; the factor of 9 should be applied once to the original 3 W.

Answer: B

Correct answer: 0.5 A. Divide voltage by current in each row: 3/0.1, 6/0.2 and 9/0.3 all equal 30 ohms, so the resistance is constant.

• A: This multiplies 15 V by the first current reading instead of dividing by the resistance.
• B (correct): Every reading gives V/I = 30 ohms, so the part is ohmic, and 15 V divided by 30 ohms is 0.5 A.
• C: This slips a decimal place; 15 divided by 30 is 0.5, not 0.05.
• D: This is the next step in the table, at 12 V; 15 V is two steps beyond 9 V, not one.

Answer: D

Correct answer: Switch off power, discharge capacitors, and lift one lead of the resistor. An ohmmeter drives its own small test current and computes R from the result, so the circuit must be off with capacitors discharged. Lifting one lead isolates the part so other paths on the board do not lower the reading.

• A: An ohmmeter supplies its own test current; outside voltage gives false readings and can damage the meter.
• B: An ohmmeter is not used in a live circuit, and a series connection measures nothing useful about R.
• C: Touching both tips puts the body in parallel with the resistor and lowers the reading.
• D (correct): The circuit must be de-energized, and lifting a lead removes parallel paths that would lower the reading.

Answer: A

Correct answer: 1 A. Two equal 12 ohm resistors in parallel act like 6 ohms, so the total is

and the source current is

The parallel pair drops 2 A × 6 ohms = 12 V, so each 12 ohm branch carries

• A (correct): The pair acts as 6 ohms, total is 11 ohms, the source current is 2 A, and each equal branch carries half.
• B: This treats all three resistors as being in series (29 ohms total).
• C: This places the full 22 V across one 12 ohm resistor, ignoring the drop across the 5 ohm resistor.
• D: This is the total source current; it splits equally between the two parallel branches.

Answer: D

Correct answer: 9 V. Series resistances add to 5.5 kilohms. In a divider each resistor takes a share proportional to its resistance.

The 2.2 kilohm resistor gets the other 6 V.

• A: This uses the full supply voltage for a resistor that gets only a share of it.
• B: This splits the voltage equally, which is only true for equal series resistors.
• C: This is the drop across the smaller 2.2 kilohm resistor.
• D (correct): The 3.3 kilohm resistor is 3.3/5.5 = 0.6 of the total resistance, so it drops 0.6 × 15 = 9 V.

Answer: A

Correct answer: 11 V. The internal resistance is in series with the load, so the total is

and the current is

The battery loses 2 A × 0.5 ohm = 1 V inside, so the terminals show

which also equals 2 A × 5.5 ohms.

• A (correct): The current is 12 ÷ 6 = 2 A; the internal drop is 2 A × 0.5 ohm = 1 V, leaving 11 V at the terminals.
• B: This is the voltage lost inside the battery, not the voltage left at its terminals.
• C: This treats the battery as ideal, ignoring the drop inside it.
• D: This computes the current from the load alone (12 ÷ 5.5), forgetting that the internal resistance is in the same loop.

Answer: C

Correct answer: More than 0.25 A, because the cooler filament has lower resistance. A tungsten filament's resistance grows as it heats. At 60 V the filament is cooler, so its resistance is below the hot value of

With less resistance, 60 V drives more than

so the lamp is nonohmic and simple proportion does not apply.

• A: Scaling with voltage holds only for constant resistance, and a filament's resistance changes with temperature.
• B: A tungsten filament's resistance falls as it cools, so the current is not reduced below the estimate.
• C (correct): At 60 V the filament runs cooler, so its resistance is below its hot value and more than half the current flows.
• D: A lamp does not force a fixed current; its current still depends on the voltage applied.

Answer: C

Correct answer: 1.5 A. Before the short, the loop has 30 ohms and carries 1 A. The short bypasses one resistor, leaving

so the current rises to

The shorted resistor's drop falls to zero.

• A: A short is a low-resistance path, not a break, so current still flows around the loop.
• B: This leaves only one 10 Ω resistor in the loop, but the short bypasses only one of the three.
• C (correct): With one resistor bypassed, the loop has 20 ohms, so the current is 30 ÷ 20 = 1.5 A.
• D: This is the current before the short; bypassing a resistor lowers the total resistance and raises the current.

Answer: A

Correct answer: 8 A. First find the hot resistance from the rating, then the cold resistance, then the current.

Because resistance rises as the filament heats, the current quickly falls toward its normal 0.5 A.

• A (correct): The hot resistance is 120² / 60 = 240 ohms, so the cold value is 15 ohms, and 120 V divided by 15 ohms is 8 A.
• B: This divides the hot current by 16 instead of multiplying, but a lower resistance gives a larger current.
• C: This stops at the cold resistance of 15 ohms and labels the number as amperes.
• D: This is the normal operating current using the hot resistance of 240 ohms, not the much lower cold resistance.

Answer: D

Correct answer: 1 A. In parallel, current divides in inverse proportion to resistance. The 60 ohm branch has twice the resistance of the 30 ohm branch, so it carries one third of the total and the 30 ohm branch carries two thirds: 1 A and 2 A.

Check: both branches share the same voltage. 1 A × 60 ohms = 60 V and 2 A × 30 ohms = 60 V, and the branch currents add to 3 A.

• A: This is the current in the 30 ohm branch; the smaller resistance takes the larger share.
• B: The 3 A total divides between the two branches, so no single branch carries all of it.
• C: This splits the current equally, but parallel branches with different resistances carry different currents.
• D (correct): The 60 ohm branch has twice the resistance of the other, so it carries one third of the 3 A total, which is 1 A.

Answer: A

Correct answer: 0.6 W. Convert first: 20 mA = 0.02 A and 1.5 kilohms = 1,500 ohms.

Check:

and 30 V × 0.02 A = 0.6 W.

• A (correct): Converting to 0.02 A and 1,500 ohms gives P = 0.0004 × 1,500 = 0.6 W.
• B: This slips a decimal place after squaring the current.
• C: This is I × R, which gives the 30 V drop across the resistor, not the power.
• D: This squares 20 and multiplies by 1.5 without converting the milli and kilo prefixes.

Answer: D

Correct answer: 2 A. The two branches share one voltage. The equivalent resistance is

so that voltage is 2.5 A × 8 ohms = 20 V. The 10 ohm branch carries

A and the 40 ohm branch carries 0.5 A, which add to 2.5 A.

• A: This is the total current; part of it flows through the 40 ohm branch.
• B: This is the current in the 40 ohm branch; the smaller resistance takes the larger share.
• C: This splits the current equally, which only happens when the branch resistances are equal.
• D (correct): The 10 ohm branch has one fourth the resistance, so it carries four fifths of 2.5 A, which is 2 A.

Answer: C

Correct answer: 0.3 A. For an ohmic part,

At 9 V the current is

and at 15 V it is 0.5 A, an increase of 0.2 A as stated.

• A: This divides 9 by the 6 V change, which mixes up the quantities.
• B: This is the increase in current, not the current at 9 V.
• C (correct): The change gives R = 6 V ÷ 0.2 A = 30 ohms, so at 9 V the current is 9 ÷ 30 = 0.3 A.
• D: This is the current at 15 V, after the increase.

Answer: C

Correct answer: 54 W. Series resistances add:

so the loop current is

A through both parts. Then use

for the 24 Ω resistor.

Check: its drop is 1.5 times 24 = 36 V, and 36 times 1.5 is also 54 W.

• A: This uses the full 48 V across the 24 Ω resistor, but in series that resistor gets only part of the supply voltage.
• B: 72 W is the total power of the circuit, 48 V times 1.5 A, not the power in one resistor.
• C (correct): The loop current is 48/32 = 1.5 A, and 1.5 squared times 24 gives 54 W.
• D: 18 W is the power in the 8 Ω resistor, not the 24 Ω one.

Answer: A

Correct answer: 0.4 A. For a fixed resistance, current is directly proportional to voltage. The voltage changes by

so the current does too: 0.6 times

Check:

• A (correct): With constant resistance, current scales with voltage: 0.6 A times 6/9 is 0.4 A.
• B: This uses the ratio upside down; lowering the voltage must lower the current.
• C: The current would stay the same only if the voltage stayed the same.
• D: This squares the voltage ratio, which is how power changes, not current.

Answer: D

Correct answer: 500 Ω, more than P's 200 Ω. Read each line at the same voltage and use

converting milliamperes to amperes first.

On a current-versus-voltage graph the slope is 1/R, so the less steep line, Q, has the larger resistance.

• A: This swaps the two resistors; the line with less current at the same voltage has more resistance.
• B: This divides volts by milliamperes without converting to amperes, so both values are 1,000 times too small.
• C: This multiplies voltage by current instead of dividing voltage by current.
• D (correct): 6 V divided by 0.012 A is 500 Ω and 6 V divided by 0.03 A is 200 Ω, so Q, the less steep line, has more resistance.

Answer: D

Correct answer: 15 mA. In series the resistances add:

Then

The same 15 mA flows through both resistors.

• A: This uses only the 470 Ω resistor; in series both resistances oppose the same current.
• B: This uses only the 330 Ω resistor and ignores the other part in the loop.
• C: This slips a decimal: 12 divided by 800 is 0.015 A, which is 15 mA.
• D (correct): Series resistances add to 800 Ω, and 12 V divided by 800 Ω is 0.015 A, or 15 mA.

Answer: A

Correct answer: 18 W. With R fixed,

so power scales with the square of the voltage. The voltage triples (

), so the power is multiplied by

2 W times 9 = 18 W. Check:

and 18 squared over 18 is 18 W.

• A (correct): Tripling the voltage at fixed R multiplies power by 3 squared, so 2 W times 9 is 18 W.
• B: This triples the power along with the voltage, but power goes with the square of the voltage.
• C: Power stays the same only if the voltage stays the same.
• D: This multiplies by 27, the cube of the voltage ratio; power depends on the square.

Answer: C

Correct answer: 3 A. Use

with the full track voltage across the motor.

Check: 3 A times 6 Ω gives back 18 V.

• A: This multiplies voltage by resistance; current comes from dividing voltage by resistance.
• B: This divides resistance by voltage, the formula turned upside down.
• C (correct): Current is voltage divided by resistance: 18 V divided by 6 Ω is 3 A.
• D: This adds volts and ohms, which are different quantities and cannot be added.

Answer: B

Correct answer: It is nonohmic, because its resistance rises as the voltage increases. Divide V by I for each row:

2/0.15 ≈ 13.3 Ω, and

Because the ratio is not constant, the part is nonohmic, and its resistance rises with voltage, the way a lamp filament behaves as it heats.

• A: Only the first reading gives 10 Ω; the V/I ratio grows at higher voltages, so R is not constant.
• B (correct): V/I goes from 10 Ω to about 13.3 Ω to 20 Ω, so the resistance climbs with voltage.
• C: The current grows more slowly than the voltage, so V/I increases rather than decreases.
• D: Rising current alone is not enough; an ohmic part needs current exactly proportional to voltage.

Answer: D

Correct answer: 0.8 A. The parallel pair is

so the total is

and the source current is

The pair's voltage is 2 A times 12 Ω = 24 V, which appears across each branch.

Check: the 20 Ω branch carries 1.2 A, and

• A: 1.2 A is the current in the 20 Ω branch; the larger 30 Ω branch carries less.
• B: This splits the 2 A total equally, but parallel branches with unequal resistance carry unequal currents.
• C: 2 A is the total source current, which divides between the two branches.
• D (correct): The pair is 12 Ω, total 18 Ω, source current 2 A, pair voltage 24 V, and 24 V / 30 Ω = 0.8 A.

Answer: B

Correct answer: 5 mA. Convert 3 kΩ to 3,000 Ω and use

Shortcut: volts divided by kilohms gives milliamperes directly.

• A: This forgets the kilo prefix; 3 kΩ is 3,000 Ω, so the current is 1,000 times smaller.
• B (correct): Volts divided by kilohms gives milliamperes: 15/3 = 5 mA, which is 0.005 A.
• C: This multiplies voltage by resistance instead of dividing.
• D: This divides resistance by voltage, the formula upside down.

Answer: A

Correct answer: 0 V. A blown fuse opens the only path, so the current is zero everywhere in the loop. With I = 0, each resistor's drop is

The full 12 V appears across the gap in the fuse instead.

• A (correct): With the loop open, no current flows, and V = IR gives zero drop across the resistor.
• B: The full 12 V appears across the gap in the blown fuse, not across a resistor that carries no current.
• C: 8 V would be this resistor's share only while current flows; with the fuse blown, current is zero.
• D: 4 V is the normal drop on the 4 Ω resistor, and it also needs current to exist.