Series Circuits Lesson Test (40 Questions)

Answer: B

Correct answer: Current has only one path from one source terminal to the other. Trace the wire from the positive terminal back to the negative terminal. If you pass through every component one after another without reaching a fork, the circuit is series.

• A: A split in the path means at least part of the circuit is parallel.
• B (correct): A single unbranched path is the defining property of a series circuit.
• C: Position on the drawing says nothing about how the parts are actually connected.
• D: Equal full voltage across each part describes parallel branches, not series.

Answer: C

Correct answer: 133 Ω. In series, the total resistance is the sum of the individual resistances.

• A: This is only the largest resistor; the series total must be larger than any single resistor.
• B: This is the average of the three values, not their sum.
• C (correct): Series resistances simply add: 12 + 39 + 82 = 133 Ω.
• D: This comes from the parallel reciprocal formula, which does not apply to series resistors.

Answer: D

Correct answer: 0.75 A. Charge cannot pile up or disappear in a series loop, so the current is identical at every point. The 30 Ω resistor carries the same 0.75 A measured at the 10 Ω resistor.

• A: This adds the current three times; current does not build up as it passes each resistor.
• B: This divides the current among the three resistors, which only happens in parallel branches.
• C: This assumes a larger resistor carries less current, but series resistors all carry the same current.
• D (correct): The same current flows through every part of a single series path.

Answer: B

Correct answer: 0.12 A. One milliampere is 0.001 A, so divide by 1,000:

• A: This divides by 10 instead of 1,000.
• B (correct): Dividing 120 mA by 1,000 gives 0.12 A.
• C: This divides by 10,000 instead of 1,000.
• D: This divides by 100 instead of 1,000.

Answer: C

Correct answer: In series, by opening the loop so the current passes through the meter. Amps go in line, volts go across. The circuit is opened at any point and the ammeter is inserted so the loop current flows through it; in a series circuit it reads the same value wherever it is placed.

• A: No current can pass through a meter that is not part of a closed path.
• B: This is how a voltmeter is connected; an ammeter there would bypass the resistor.
• C (correct): An ammeter measures the current flowing through it, so it must be placed in line with the circuit.
• D: An ammeter across the source acts as a near short and can blow its fuse or be damaged.

Answer: A

Correct answer: 5.5 V. The drops in a series loop sum to the source voltage. Subtract the known drops:

• A (correct): The drops must add to 18 V, so 18 − 12.5 = 5.5 V.
• B: This adds the known drops to the source; no single drop can exceed the source voltage.
• C: This splits 18 V evenly three ways, which is only true for identical resistors.
• D: This is the sum of the two known drops, not what remains of the source voltage.

Answer: D

Correct answer: 12 V. Apply Ohm's law to the resistor using its own values.

• A: This divides resistance by current instead of multiplying them.
• B: This adds amperes and ohms as if they were the same kind of quantity.
• C: This divides current by resistance instead of multiplying them.
• D (correct): Ohm's law gives 0.3 A × 40 Ω = 12 V.

Answer: D

Correct answer: All the bulbs go dark. A burned-out filament is an open in the circuit. Since a series circuit has only one path, any single open stops the current through every component, and all the bulbs go out.

• A: Current stops everywhere in the loop at once, not just downstream of the break.
• B: That would happen in a parallel string; in series the open stops current everywhere.
• C: Brighter bulbs would need more current, but an open drops the current to zero.
• D (correct): The broken filament opens the only path, so current stops through every bulb.

Answer: C

Correct answer: Across the resistor, in parallel with it. A voltmeter measures the potential difference between two points, so its leads touch both ends of the resistor, side by side with it. Its very high resistance means it draws almost no current.

• A: The meter must touch both ends of the resistor to measure the difference between them.
• B: That is how an ammeter is connected; a high-resistance voltmeter in line would nearly stop the current.
• C (correct): A voltmeter compares the two ends of the part, so its leads go across it without breaking the circuit.
• D: That only gives the right value for identical resistors and does not measure the drop directly.

Answer: A

Correct answer: 0 V. A closed ideal switch has zero resistance. With current I flowing, its drop is

and the entire 12 V is shared by the two resistors.

• A (correct): An ideal closed switch acts like a plain wire with zero resistance, so it has no voltage drop.
• B: This splits the source among three parts, but a closed switch adds no resistance and takes no share.
• C: The full source voltage appears across a switch only when it is open.
• D: This splits the source between the switch and the resistors as if the switch had resistance.

Answer: D

Correct answer: 0.6 A. Add the resistances, then use Ohm's law for the whole loop.

• A: This uses the parallel combination of 7.5 Ω instead of adding the resistors.
• B: This uses the full source voltage across only the 10 Ω resistor.
• C: This uses the full source voltage across only the 30 Ω resistor.
• D (correct): The total resistance is 40 Ω, and 24 V ÷ 40 Ω = 0.6 A.

Answer: C

Correct answer: 35 V. Use the voltage-divider relationship with a total of 90 Ω.

Check: the 20 Ω resistor drops 10 V, and

• A: This is the drop across the 20 Ω resistor, not the 70 Ω resistor.
• B: A series resistor shares the source voltage; it does not receive all of it.
• C (correct): The 70 Ω resistor is 70/90 of the total, and 45 × 70/90 = 35 V.
• D: This splits the voltage evenly, which only applies to equal resistors.

Answer: C

Correct answer: 3 mA. Add the resistances: 1.8 kΩ + 2.2 kΩ = 4 kΩ = 4,000 Ω. Then

which is 3 mA. The shortcut volts divided by kilohms gives milliamperes leads to the same result.

• A: This divides 12 by 4 without converting kilohms to ohms, giving a value 1,000 times too large.
• B: This misplaces the decimal; 0.003 A equals 3 mA, not 0.3 mA.
• C (correct): The total is 4 kΩ, or 4,000 Ω, and 12 ÷ 4,000 = 0.003 A = 3 mA.
• D: This is ten times too large; 12 V ÷ 4 kΩ is 3 mA.

Answer: B

Correct answer: 17 V. Convert first: 25 mA = 0.025 A.

• A: This converts 25 mA to 0.0025 A, dividing by 10,000 instead of 1,000.
• B (correct): Converting 25 mA to 0.025 A and multiplying by 680 Ω gives 17 V.
• C: This divides 680 by 25 instead of multiplying current by resistance.
• D: This multiplies 25 by 680 without converting milliamperes to amperes.

Answer: B

Correct answer: 20 V. Find the loop current from the known drop:

The total resistance is

so the source is

Check: the 30 Ω drops 12 V, and

• A: This adds 8 volts to 30 ohms, mixing two different quantities.
• B (correct): The current is 8 ÷ 20 = 0.4 A, and 0.4 A × 50 Ω = 20 V.
• C: This assumes the source equals one resistor's drop, ignoring the other resistor.
• D: This is only the drop across the 30 Ω resistor; the source supplies both drops.

Answer: A

Correct answer: 36 Ω. In series, the ratio of drops equals the ratio of resistances.

• A (correct): The current is 3 ÷ 12 = 0.25 A, and 9 V ÷ 0.25 A = 36 Ω.
• B: This inverts the ratio; the resistor with the larger drop must have the larger resistance.
• C: This multiplies 12 by 9 without dividing by the 3 V drop of the known resistor.
• D: This multiplies the drop by the current instead of dividing the drop by the current.

Answer: A

Correct answer: 50 W. Because the current is common in series, use

• A (correct): Using P = I²R: 2.5² × 8 = 6.25 × 8 = 50 W.
• B: This divides resistance by current, which is not a power formula.
• C: This squares the resistance instead of the current.
• D: This multiplies current by resistance, which gives the voltage drop in volts, not power.

Answer: B

Correct answer: 15 V. Opening the switch stops the current everywhere, so each resistor drops

By the loop rule, the drops must still total 15 V, so the full 15 V appears across the open switch.

• A: Zero is what each resistor reads after the switch opens, not the open gap.
• B (correct): With no current, both resistors drop 0 V, so the entire source voltage appears across the open contacts.
• C: This is the 2 Ω resistor's drop while current was flowing, before the switch opened.
• D: This splits the source evenly between the gap and the resistors, but with no current the resistors drop nothing.

Answer: A

Correct answer: 6 A. Before the bypass,

and I = 2 A. The wire removes the 8 Ω resistor from the path, so

• A (correct): The bypassed resistor drops out of the total, leaving 24 V ÷ 4 Ω = 6 A.
• B: This divides the source by the bypassed 8 Ω resistor, which no longer carries current.
• C: This treats the wire as an open; a bypass gives current an easier path and raises it.
• D: This is the original current of 24 V ÷ 12 Ω, before the bypass.

Answer: B

Correct answer: It falls from 30 V to 20 V. Alone, the 10 Ω resistor drops the full 30 V. With 5 Ω added,

so its drop becomes

V and the 5 Ω resistor takes the other 10 V.

• A: Adding resistance lowers the current, so the original resistor's drop must fall.
• B (correct): The new current is 30 ÷ 15 = 2 A, so the 10 Ω resistor drops 2 × 10 = 20 V.
• C: This splits the source in half, which would require two equal resistors.
• D: No drop in the loop can exceed the 30 V source.

Answer: A

Correct answer: 9 V. Replacing a lamp with a wire lowers the total resistance but keeps the path closed. With two identical lamps left, each gets

up from 6 V, so they glow brighter.

• A (correct): Two identical lamps now share 18 V equally, 9 V each.
• B: Each lamp gets the full source only if it is the only load; two lamps share it.
• C: This treats the wire as an open; replacing a lamp with a wire keeps the loop complete.
• D: This is each lamp's share before the change, when three lamps divided the voltage.

Answer: C

Correct answer: The 5 kΩ resistor. With a common current, each drop

is proportional to resistance. The 5 kΩ resistor takes 5/10, or half, of the source voltage, more than either of the others.

• A: Its position in the loop does not matter; only its resistance relative to the others does.
• B: Equal current does not mean equal drops; drops are proportional to resistance.
• C (correct): Every resistor carries the same current, so the largest resistance has the largest IR drop.
• D: This reverses the rule; the smallest resistor takes the smallest share of the voltage.

Answer: B

Correct answer: 2 Ω: 2 V; 4 Ω: 4 V; 10 Ω: 10 V. Total resistance is

The drops are

V and

which add to 16 V.

• A: This gives every resistor the full source voltage, which belongs to parallel circuits.
• B (correct): The current is 16 ÷ 16 = 1 A, and each drop is 1 A times its resistance.
• C: This gives the biggest drop to the smallest resistor, which reverses the series rule.
• D: These sum to 16 V but are not proportional to the resistances, so the current would differ between resistors.

Answer: B

Correct answer: 10 V. Four equal resistors on 20 V drop 5 V each. Moving from the positive terminal, the node voltages fall 20 V, 15 V, 10 V, 5 V and 0 V, so the node after the second resistor sits at 10 V.

• A: This is the drop across a single resistor, not the node voltage after two drops.
• B (correct): Each resistor drops 5 V, so after two resistors the node is at 20 − 10 = 10 V.
• C: This is the voltage at the positive terminal, before any resistor.
• D: This subtracts only one 5 V drop, which is the node between the first and second resistors.

Answer: D

Correct answer: Part of the voltage drops across the battery's own internal resistance. A real battery behaves like an ideal source in series with a small internal resistance. Under load, current flows through that resistance and drops some voltage inside the battery, so the terminal voltage falls below the no-load value.

• A: Current is the same all around a series loop; it is not used up by loads.
• B: The reading went down, and a heavier load lowers terminal voltage rather than raising it.
• C: An ideal voltmeter draws almost no current, so it does not explain the drop.
• D (correct): A real battery acts like an ideal source in series with a small internal resistance that drops voltage when current flows.

Answer: A

Correct answer: So that when it melts, it opens the only path and stops all the current. A fuse is a thin metal link in the circuit's path. Under normal current it acts like a wire; if current exceeds its rating, the link melts and creates an open, stopping current everywhere in that series path.

• A (correct): A series fuse carries the full circuit current, and melting it breaks the path for every component.
• B: Current does not divide in series; a fuse in series carries the same current as the load.
• C: A fuse acts like a wire under normal current; it does not increase the load voltage.
• D: A series fuse carries the circuit current at all times, not just during a fault.

Answer: A

Correct answer: 36 W. With a common current, each resistor's fraction of the total power equals its fraction of the total resistance. The 18 Ω resistor is 18 of 24 Ω, so it dissipates

leaving 12 W for the 6 Ω resistor.

• A (correct): The 18 Ω resistor is 18/24 of the total resistance, and 48 × 18/24 = 36 W.
• B: This splits the power evenly, but series power divides in proportion to resistance.
• C: This is the total power for both resistors, not one resistor's share.
• D: This is the 6 Ω resistor's share, not the 18 Ω resistor's share.

Answer: C

Correct answer: The current falls and the lamp gets less voltage. The bad connection is an extra resistance in series. Total resistance increases, so the loop current falls, and the clip drops some voltage itself, leaving less for the lamp. The bad connection may also get hot.

• A: The current is equal at every point, but its value drops when total resistance increases.
• B: The clip takes its own drop, so the lamp's share of the voltage shrinks.
• C (correct): The extra series resistance raises the total, lowering the current and taking part of the voltage away from the lamp.
• D: Adding resistance in series lowers current; it cannot raise it.

Answer: C

Correct answer: 25 V. Total resistance is

A voltmeter across two series resistors reads the sum of their drops: 10 V + 15 V = 25 V.

• A: This is the drop across the 15 Ω resistor alone, not across both resistors.
• B: This is the drop across the 10 Ω resistor alone, not across both resistors.
• C (correct): The current is 1 A, so the two resistors drop 10 V and 15 V, which total 25 V.
• D: This is the full source voltage, which would include the 5 Ω resistor's 5 V drop.

Answer: B

Correct answer: The loop has a hidden branch or a measurement error. Charge conservation means the current is identical everywhere in one unbranched path. Unequal readings prove the two points are not in a single series path, or that a meter or scale was misread.

• A: Current stays the same all around a series loop even as voltage drops.
• B (correct): In a true series path every ammeter reading must match, so unequal readings point to a branch or a meter problem.
• C: A low source lowers the single current value, but it cannot make two readings differ in one path.
• D: Resistors reduce voltage, not current; charge is not consumed in a resistor.

Answer: A

Correct answer: 15 V. Add the resistances, find the current, then apply Ohm's law to the 10 Ω resistor.

Check: 9 V + 15 V + 24 V = 48 V.

• A (correct): The total is 32 Ω, so I = 1.5 A, and 1.5 A × 10 Ω = 15 V.
• B: This splits 48 V evenly three ways, which only works for identical resistors.
• C: This finds current from the full 48 V on the 6 Ω resistor alone; no drop can exceed the source.
• D: This is the drop across the 16 Ω resistor, not the 10 Ω resistor.

Answer: C

Correct answer: 96 W. Before:

I = 2 A, and the 6 Ω resistor uses

After the bypass:

• A: This treats the bypass as an open; only the bypassed 9 Ω resistor stops carrying current.
• B: This is the 6 Ω resistor's power before the bypass, when the current was 2 A.
• C (correct): The new total is 9 Ω, so I = 4 A and P = 4² × 6 = 96 W.
• D: This doubles the power because the current doubled, but power depends on current squared.

Answer: D

Correct answer: 50 V. Solve

and I = 0.5 A. The total resistance is 100 Ω, so the source is

Check: the drops are 20 V and 30 V, totaling 50 V.

• A: This is only the drop across the 60 Ω resistor, not the whole source.
• B: This is only the drop across the 40 Ω resistor, not the whole source.
• C: This uses I² = 0.25 as the current, forgetting to take the square root.
• D (correct): From I² × 40 = 10, I = 0.5 A, and 0.5 A × 100 Ω = 50 V.

Answer: D

Correct answer: 2.4 kΩ. Convert 2.5 mA to 0.0025 A. Then

or 3.6 kΩ. Subtract the known resistor: 3.6 kΩ − 1.2 kΩ = 2.4 kΩ.

• A: The arithmetic is right but the kilohm unit was dropped, making it 1,000 times too small.
• B: This adds the known resistor to the total instead of subtracting it.
• C: This is the total resistance; the known 1.2 kΩ must still be subtracted.
• D (correct): The total is 9 ÷ 0.0025 = 3,600 Ω, and 3.6 kΩ − 1.2 kΩ = 2.4 kΩ.

Answer: C

Correct answer: 5 V. Before the change each 5 Ω resistor dropped 10 V. After it,

so the 5 Ω resistor drops

V and the 15 Ω resistor takes 15 V.

• A: This is the drop across the new 15 Ω resistor, not the 5 Ω resistor.
• B: This is the drop before the change; the larger resistor lowers the current and this drop.
• C (correct): The new total is 20 Ω, so I = 1 A and the 5 Ω resistor drops 5 V.
• D: This gives the 5 Ω resistor the full source voltage, ignoring the other resistor.

Answer: D

Correct answer: The 12 Ω lamp, at 12 W. Total resistance is 16 Ω, so

A through both lamps. Using

the 4 Ω lamp uses 4 W and the 12 Ω lamp uses 12 W, so the higher-resistance lamp dissipates more. Check:

W = 16 V × 1 A.

• A: This uses the full 16 V across the 12 Ω lamp instead of its own 12 V drop.
• B: This applies the full 16 V to the 4 Ω lamp alone instead of its own 4 V drop.
• C: Equal current does not mean equal power; power is proportional to resistance in series.
• D (correct): The current is 1 A, so the 12 Ω lamp dissipates 1² × 12 = 12 W versus 4 W for the other.

Answer: A

Correct answer: It falls below 6 V. A voltage divider assumes no current is drawn from its midpoint. A load connected across the lower resistor is in parallel with it, so the lower section's effective resistance falls, its fraction of the total falls, and the output drops below the unloaded 6 V.

• A (correct): The motor sits in parallel with the lower resistor, lowering that section's effective resistance and its share of the voltage.
• B: The divider formula assumes nothing draws current from the midpoint, which is no longer true.
• C: The motor is across only the lower resistor, and the upper resistor still limits the current.
• D: A load across one resistor is in parallel with it, which lowers the resistance there rather than adding to it.

Answer: D

Correct answer: The meter bypasses the 10 Ω resistor and the current rises from 1 A to 6 A. Normally

An ideal ammeter has nearly zero resistance, so placing it across the 10 Ω resistor shorts that resistor out. The loop is left with 2 Ω, and the current jumps to

which in a real circuit could blow the meter's fuse.

• A: An ammeter measures current, not voltage, and placing it across the resistor changes the circuit.
• B: Connecting a meter across a part does not break the path; a low-resistance meter adds an easier path.
• C: That reading requires inserting the meter in series, not placing it across a resistor.
• D (correct): The near-zero-resistance meter acts as a short, leaving only 2 Ω in the loop, so 12 ÷ 2 = 6 A flows.

Answer: D

Correct answer: The third lamp is open. When a series loop is open, the current is zero, so every intact component drops 0 V. By the loop rule the entire 12 V appears across the break, which places the open at the third lamp.

• A: A shorted lamp would show 0 V, and the other two lamps would light with 6 V each.
• B: An open first lamp would show 12 V across the first lamp and 0 V across the others.
• C: A dead battery would give 0 V across every lamp, not 12 V across one of them.
• D (correct): With no current, intact lamps drop 0 V, and the full source voltage appears across the open part.

Answer: B

Correct answer: 4 Ω. The source delivers

so the third resistor dissipates

Check: the resistors are 3 Ω, 5 Ω and 4 Ω, totaling 12 Ω, and

• A: This is the total resistance of the loop, 24 V ÷ 2 A, not the third resistor alone.
• B (correct): The source supplies 48 W, so the third resistor uses 16 W, and 16 ÷ 2² = 4 Ω.
• C: This divides 16 W by the current, 2 A, instead of by the current squared.
• D: This divides the current squared by the power, inverting the formula.

Answer: A

Correct answer: 188 Ω. In series,

Four equal 47 Ω parts give

larger than any single resistor.

• A (correct): Series resistances add: 47 times 4 is 188 Ω.
• B: This divides by 4, which is the rule for equal resistors in parallel, not series.
• C: This treats the string as one resistor, but each one adds its own resistance.
• D: This adds the number of resistors to one value instead of adding all four values.

Answer: A

Correct answer: 10 mA. Add the resistances:

which is 10 mA through both resistors.

• A (correct): The total resistance is 900 Ω, and 9 V divided by 900 Ω is 0.01 A, or 10 mA.
• B: This uses only the 220 Ω resistor instead of the total series resistance.
• C: This uses only the 680 Ω resistor instead of the total series resistance.
• D: This divides 9 by 90, a decimal slip when dividing by 900.

Answer: B

Correct answer: 2 kΩ. The known drop gives the loop current:

or 2.5 mA.

Total resistance:

Subtract the known parts:

or 2 kΩ.

• A: This is the total resistance of the whole loop, not the unknown resistor.
• B (correct): The current is 4.5 V over 1.8 kΩ, or 2.5 mA, so the total is 6 kΩ and R is 6 minus 4, or 2 kΩ.
• C: This subtracts only the 1.8 kΩ resistor from the total and forgets the 2.2 kΩ resistor.
• D: This has the right digits but loses the kilo prefix in the unit conversion.

Answer: B

Correct answer: 13.5 W. After the change,

Lamp power:

down from 24 W before the resistor was added.

• A: This is the lamp power before the resistor was added; the added resistance lowers the current.
• B (correct): The new current is 12 over 8, or 1.5 A, and $1.5^2\times 6=13.5$ W.
• C: This is the total power of the loop, which includes the 2 Ω resistor.
• D: This is the power in the added 2 Ω resistor, not in the lamp.

Answer: A

Correct answer: 6 V. Use the divider rule:

The 3 kΩ resistor drops the other 18 V.

• A (correct): The 1 kΩ resistor is one-fourth of the 4 kΩ total, so it takes one-fourth of 24 V.
• B: This is the drop across the 3 kΩ resistor, the larger share.
• C: This divides 24 V by 3, using the other resistor value instead of the total.
• D: This gives one series resistor the full source voltage, which only happens when it is alone.

Answer: C

Correct answer: It decreases. In series, $R_T$ grows with every resistor added. Since

and the source voltage is fixed, a larger total resistance means a smaller current.

• A: This confuses series with parallel, where adding a path raises the total current.
• B: The current depends on the total resistance, which has changed.
• C (correct): The added resistor raises the total resistance, so the same voltage pushes less current.
• D: Adding a resistor does not break the loop, so current still flows.

Answer: C

Correct answer: 20 V. Find the shared current from the known drop:

Total resistance is

Check: drops of 4 V, 6 V and 10 V add to 20 V.

• A: This assumes every resistor drops 4 V, but unequal resistors have unequal drops.
• B: This multiplies the 4 V drop by the total resistance instead of using the current.
• C (correct): The current is 4 V over 8 Ω, or 0.5 A, and 0.5 A times the 40 Ω total is 20 V.
• D: This is only the drop across the 20 Ω resistor, not the whole source.

Answer: D

Correct answer: 0.6 W. Convert first: 20 mA is 0.02 A and 1.5 kΩ is 1,500 Ω.

Then

• A: This squares 20 instead of 0.02, skipping the milliampere conversion.
• B: This is current times resistance, which gives the 30 V drop, not power.
• C: This leaves the resistance as 1.5 instead of converting kilohms to 1,500 Ω.
• D (correct): Converting to 0.02 A and 1,500 Ω gives $0.0004\times 1500=0.6$ W.

Answer: D

Correct answer: 6 V. Total resistance is 12 Ω, so the current is

The drops are 2 V, 4 V and 6 V.

Starting at 12 V and subtracting the first two drops leaves

which equals the 6 V still to be dropped across the last resistor.

• A: This is the drop across the 4 Ω resistor, not the node voltage.
• B: This is the node voltage after only the 2 Ω resistor.
• C: This subtracts only the 4 Ω drop and forgets the drop across the 2 Ω resistor.
• D (correct): The 2 Ω and 4 Ω resistors drop 2 V and 4 V, so 12 minus 6 leaves 6 V at that node.

Answer: C

Correct answer: 12 V. Think in shares of resistance. With three resistors of R and one of 2R, the total is 5R.

The larger resistor gets

and each smaller one gets 6 V. Check:

• A: This is each resistor’s share before the change, one-fourth of 30 V.
• B: This doubles the old 7.5 V share without allowing for the larger total resistance.
• C (correct): The shares are now 1, 1, 1 and 2 out of 5, so the larger resistor takes 2/5 of 30 V.
• D: This is the new drop across each of the three smaller resistors.