Parallel Circuits Lesson Test: 40 EI Questions

Answer: A

Correct answer: Each end of one resistor connects to the same node as an end of the other. Trace the wire from each terminal. If both resistors connect between the same two nodes, with no component in between, they are in parallel no matter how they are drawn or what their values are.

• A (correct): Parallel means both resistors span the same pair of nodes, so they share one voltage.
• B: Equal values say nothing about how the parts are wired; unequal resistors can be parallel and equal ones can be in series.
• C: One current flowing through parts in turn describes a series connection, not a parallel one.
• D: Drawing position does not matter; only the nodes each terminal connects to decide the connection.

Answer: A

Correct answer: 18 V. In a simple parallel group, each branch spans the same two nodes as the source. So the 6 Ω resistor has the full 18 V across it, the same as the 3 Ω and 9 Ω resistors.

• A (correct): Every branch connected across the same two rails has the full source voltage across it.
• B: Branch voltages in parallel do not add up; multiplying 18 V by three branches gives a value larger than the source can supply.
• C: 18 divided by 6 gives the branch current in amperes, not the voltage across the branch.
• D: Dividing the 18 V equally among three resistors treats the circuit as if voltage were split, which happens in series, not parallel.

Answer: D

Correct answer: 3 A. Each parallel branch has the supply voltage across it, so use Ohm's law on the coil alone.

• A: Dividing the resistance by the voltage inverts Ohm's law.
• B: Halving the result assumes the coil shares the supply voltage with another branch, but parallel branches each get the full voltage.
• C: Multiplying voltage by resistance does not give current; Ohm's law divides the voltage by the resistance.
• D (correct): The branch has the full 24 V across it, and 24 V divided by 8 Ω is 3 A.

Answer: B

Correct answer: 5 Ω. For identical resistors in parallel, divide one value by the number of branches: 15 Ω divided by 3 gives 5 Ω, which is below 15 Ω as every parallel result must be.

• A: Adding the three values is the series rule; a parallel equivalent must be smaller than any one branch.
• B (correct): For n equal resistors in parallel, the equivalent is R divided by n, and 15 divided by 3 is 5.
• C: Adding equal branches does not keep the resistance the same; each extra path lowers the total.
• D: Dividing by 2 is the rule for two equal resistors, but here there are three.

Answer: D

Correct answer: 3.5 A. By Kirchhoff's current law, current in equals current out at the junction. So the second branch carries 9 A minus 5.5 A, which is 3.5 A.

• A: Splitting the 9 A in half ignores the stated 5.5 A in the first branch.
• B: The full 9 A cannot go through the second branch when 5.5 A already leaves through the first.
• C: Adding the two numbers makes the outgoing current larger than the incoming current, which would create charge at the junction.
• D (correct): The currents leaving the junction must total the 9 A coming in, and 9 minus 5.5 is 3.5.

Answer: C

Correct answer: Open that branch and insert the meter so the branch current passes through it. An ammeter reads the current flowing through itself, so it must be in series with the path of interest. To read one branch, break that branch and insert the meter there; the main line would give the total instead.

• A: The main line carries the total of all branch currents, not the current of the one branch.
• B: An ammeter across the source shorts the supply rather than measuring any branch current.
• C (correct): An ammeter must be in series with the path it measures, so the branch is broken and the meter completes it.
• D: Placing a low-resistance ammeter across a branch is effectively a short across the rails and can blow the meter's fuse.

Answer: D

Correct answer: They stay lit at the same brightness. A burned-out filament is an open in one branch, so only that branch stops carrying current. The other branches keep the full supply voltage, so their current and brightness are unchanged; only the total current from the source drops.

• A: The voltage across each remaining branch does not rise, so its current and power do not increase.
• B: The total current drops only because one branch is gone; each remaining branch keeps the same current.
• C: Every lamp going dark is series behavior; in parallel the open stops only its own branch.
• D (correct): Each remaining lamp still connects across the same two rails, so its voltage, current and power are unchanged.

Answer: C

Correct answer: 8 Ω. Use product over sum for two resistors.

The result is below 12 Ω, as required.

• A: Adding 1/12 and 1/24 gives 1/8, but the final reciprocal was skipped, leaving a value in siemens.
• B: Averaging the two resistances does not give a parallel equivalent and is larger than the 12 Ω branch.
• C (correct): Product over sum gives 288 divided by 36, which is 8 Ω, smaller than the 12 Ω branch.
• D: Adding the two values is the series rule; a parallel result must be below the smaller resistor.

Answer: B

Correct answer: 24 W. In a parallel branch the voltage is known, so use

144 divided by 6 gives 24 W. Check with

the current is 2 A, and 12 V times 2 A is 24 W.

• A: Multiplying voltage by resistance is not a power formula; power uses voltage squared divided by resistance.
• B (correct): With the full 12 V across it, the power is 12 squared, or 144, divided by 6, which is 24 W.
• C: 2 is the branch current in amperes; it still has to be multiplied by the 12 V to get power.
• D: Squaring the voltage is only the first step; the result must also be divided by the 6 Ω.

Answer: D

Correct answer: 3 mA. Divide the branch voltage by its resistance: 9 V divided by 3,000 Ω is 0.003 A. Using the shortcut, volts over kilohms gives milliamperes, so the current is 3 mA.

• A: Multiplying the voltage by the resistance is not Ohm's law for current.
• B: Treating kilohms as ohms gives an answer a thousand times too large.
• C: Dividing the resistance by the voltage inverts the relationship.
• D (correct): Volts divided by kilohms gives milliamperes directly, and 9 divided by 3 is 3 mA.

Answer: B

Correct answer: 5 A. Each branch has the full 30 V. The 10 Ω branch carries 30 divided by 10, or 3 A, and the 15 Ω branch carries 30 divided by 15, or 2 A. The source supplies their sum, 5 A.

• A: Dividing 30 V by 25 Ω treats the two resistors as if they were in series.
• B (correct): The branches carry 3 A and 2 A, and those add to 5 A at the source.
• C: 2 A is only the current in the 15 Ω branch; the source must supply both branches.
• D: 3 A is only the current in the 10 Ω branch; the 15 Ω branch adds its own current.

Answer: C

Correct answer: 4 Ω. Use a common denominator of 24.

Flip the result:

which is below the smallest branch of 8 Ω.

• A: This is the sum of the reciprocals, 1/4, left unflipped; it is a conductance in siemens.
• B: Adding the three resistances is the series rule, giving a value larger than any branch.
• C (correct): The reciprocals add to 6/24, which is 1/4, and flipping that gives 4 Ω.
• D: Multiplying all three values and dividing by their sum is a misuse of product over sum, which only works for two resistors at a time.

Answer: C

Correct answer: 9 Ω. The known branch carries 36 divided by 12, or 3 A. By the junction rule, the other branch carries 7 A minus 3 A, or 4 A. Its resistance is 36 V divided by 4 A, which is 9 Ω.

• A: Adding the 3 A to the 7 A instead of subtracting it gives 10 A and the wrong result.
• B: 36 V divided by 7 A gives the equivalent resistance of the whole network, not one branch.
• C (correct): The 12 Ω branch takes 3 A, leaving 4 A for the other branch, and 36 V divided by 4 A is 9 Ω.
• D: 4 is the unknown branch's current in amperes; it still has to be divided into the 36 V.

Answer: D

Correct answer: 6 A. For two branches, the current in one is the total times the other branch's resistance over the sum: 9 A times 12/18 gives 6 A. Check: the 12 Ω branch gets 3 A, and both branches have 36 V across them.

• A: Splitting the current equally ignores that the smaller resistance carries more current.
• B: 3 A is the current in the 12 Ω branch; the ratio was applied backward.
• C: The full 9 A is the total; part of it goes through the 12 Ω branch.
• D (correct): The currents split in the inverse ratio 12 to 6, or 2 to 1, so the 6 Ω branch gets two thirds of 9 A.

Answer: C

Correct answer: 5 A. Adding a parallel branch does not change the voltage on the existing branch, so the 6 Ω resistor still draws 12 divided by 6, or 2 A. The new branch draws 12 divided by 4, or 3 A, so the source now supplies 5 A.

• A: Leaving the total at 2 A ignores the current the new path adds.
• B: 12 V divided by 10 Ω treats the added resistor as if it were in series.
• C (correct): The 6 Ω branch still carries 2 A and the new 4 Ω branch adds 3 A, for 5 A in total.
• D: 3 A is only the new branch's current; the original branch keeps drawing its 2 A.

Answer: C

Correct answer: 8 A. Before the fault, the branch currents were 6 A, 4 A and 2 A, for 12 A. When the 6 Ω branch opens, only its 4 A disappears; the other branches keep their 24 V, so the total becomes 6 A plus 2 A, or 8 A.

• A: 4 A is the current the open branch used to carry, not the current that remains.
• B: 12 A was the total before the fault; the open branch no longer contributes its 4 A.
• C (correct): The 4 Ω and 12 Ω branches still carry 6 A and 2 A, which add to 8 A.
• D: An open in one branch does not stop the other parallel branches from conducting.

Answer: A

Correct answer: 48 W. Each branch has 12 V, so

W and

a total of 48 W. Check: the currents are 3 A and 1 A, and 12 V times 4 A is 48 W.

• A (correct): The branches dissipate 36 W and 12 W, and the source delivers their sum.
• B: 36 W is the power in the 4 Ω branch only; the 12 Ω branch adds 12 W.
• C: 144 divided by 16 Ω treats the resistors as if they were in series.
• D: 4 is the total current in amperes; it must be multiplied by 12 V to give power.

Answer: D

Correct answer: 2 A. Reduce in stages. Two equal 12 Ω resistors in parallel act as 6 Ω. Add the 3 Ω series resistor for 9 Ω total, so the source current is 18 V divided by 9 Ω, or 2 A.

• A: Adding all three resistors to get 27 Ω treats the parallel pair as if it were in series.
• B: 1 A is the current in each 12 Ω branch, which is half of the source current.
• C: Dividing 18 V by only the 3 Ω series resistor ignores the parallel block.
• D (correct): The parallel pair is 6 Ω, the total is 9 Ω, and 18 V divided by 9 Ω is 2 A.

Answer: B

Correct answer: 12 V. The parallel pair is 6 times 3 over 9, or 2 Ω, so the total is 4 Ω and the source current is 24 divided by 4, or 6 A. The 2 Ω resistor drops 12 V, leaving 24 minus 12, or 12 V, across each resistor in the parallel group.

• A: Subtracting the 6 A current from 24 V mixes amperes with volts.
• B (correct): The source current is 6 A, the series resistor drops 12 V, and the remaining 12 V appears across both parallel resistors.
• C: The full supply voltage does not reach the parallel group because the series resistor uses some of it first.
• D: Splitting the group voltage between the two resistors treats them as series; parallel members share one voltage.

Answer: C

Correct answer: It is slightly less than 100 Ω. Every new parallel path adds its own current at the same voltage, which lowers the equivalent resistance. A huge resistor adds only a trickle of current, so the result is just under 100 Ω.

• A: Adding any positive resistor in parallel can only lower the equivalent, never raise it.
• B: Even a very large branch carries a small current, so the equivalent changes by a small amount.
• C (correct): The large resistor adds a tiny extra current path, so the equivalent drops just below 100 Ω.
• D: A parallel equivalent is always below the smallest branch, which here is 100 Ω.

Answer: C

Correct answer: 0.35 S. Conductance is the reciprocal of resistance, and parallel conductances add:

S. Its reciprocal, about 2.86 Ω, would be the equivalent resistance.

• A: Taking the reciprocal of 0.35 gives the equivalent resistance in ohms, not the conductance in siemens.
• B: Adding the resistances does not give conductance; conductance is the reciprocal of each resistance.
• C (correct): The conductances are 0.2 S, 0.1 S and 0.05 S, and in parallel they add to 0.35 S.
• D: 0.05 S is the conductance of the 20 Ω branch alone; the other branches add theirs.

Answer: A

Correct answer: The battery is shorted, and current becomes very large until something limits it. In a simple parallel group, every resistor spans the same two rails, so a wire across one of them connects the rails directly. The equivalent resistance becomes essentially zero, and the current is limited only by internal resistance, wiring, a fuse or a breaker.

• A (correct): A wire across one parallel resistor also lies across every branch and the source, so it is a short of the supply.
• B: A near-zero resistance path draws a very large current, not zero current.
• C: The rail voltage collapses toward zero in a real short, so the remaining resistors get almost no current.
• D: That would be true in a series chain, but here the wire spans the same rails as every branch and the source.

Answer: B

Correct answer: 18 V. Branch A has the full 24 V across it and a resistance of

so it carries

Inside that branch, the 6 Ω resistor drops 3 A times 6 Ω, or 18 V, and the 2 Ω drops the remaining 6 V.

• A: 6 V is the drop across the 2 Ω resistor, not the 6 Ω resistor.
• B (correct): Branch A totals 8 Ω and carries 3 A, so the 6 Ω resistor drops 3 A times 6 Ω, or 18 V.
• C: The 24 V is across the whole branch A; it is split between the two series resistors in that branch.
• D: Splitting the branch voltage evenly ignores that the larger resistor in a series pair takes the larger share.

Answer: B

Correct answer: All three lamps go out. An open switch acts like a break. Placed in the main line before the first junction, it cuts the feed to every branch at once, so all lamps go dark. A switch inside one branch would control only that branch.

• A: The switch is not inside any one branch, so it does not control a single lamp.
• B (correct): A switch ahead of the first junction breaks the path that feeds every branch.
• C: Opening the main line removes the supply from every branch, so no lamp can gain power.
• D: Parallel branches are independent of each other, but all of them still depend on the shared main line.

Answer: C

Correct answer: 4 A. Parallel branches share one voltage: 0.8 A times 15 Ω is 12 V. The 3 Ω branch then carries 12 divided by 3, or 4 A. This matches the inverse ratio: one fifth the resistance gives five times the current.

• A: Equal currents in unequal parallel branches would require different voltages, which parallel branches cannot have.
• B: 4.8 A is the total of both branch currents, not the current in the 3 Ω branch.
• C (correct): The shared voltage is 0.8 A times 15 Ω, or 12 V, and 12 V divided by 3 Ω is 4 A.
• D: Applying the ratio backward gives the smaller resistor less current, but in parallel it carries more.

Answer: C

Correct answer: The other parallel branches give the meter's test current extra paths. An ohmmeter measures everything connected between its two leads. Other branches across the same nodes act as parallel paths, so the reading is the equivalent resistance, which is lower. Lifting one end of the resistor gives its true value.

• A: A resistor's value does not change because it is installed; the surrounding paths change the reading.
• B: An ohmmeter supplies its own test current and must be used on an unpowered circuit.
• C (correct): The meter sees the resistor in parallel with the rest of the network, which lowers the reading.
• D: An ohmmeter is placed across the part; the fix is to disconnect one end of the resistor from the network.

Answer: B

Correct answer: 360 W. For a fixed resistance,

The voltage ratio is

so the power ratio is 2 squared, or 4, and 90 W times 4 is 360 W. (The element's resistance is

W confirms it.)

• A: Dividing by four reverses the effect; a higher voltage raises the power.
• B (correct): Power depends on voltage squared, so doubling the voltage multiplies the power by four.
• C: Doubling the power assumes power is proportional to voltage, but it varies with the square of voltage.
• D: The power cannot stay the same when the voltage across a fixed resistance changes.

Answer: B

Correct answer: 15 Ω. Work backward with reciprocals.

So R is 15 Ω. Check: 10 times 15 over 25 is 6 Ω.

• A: Subtracting 6 from 10 treats the resistances as if they added in series.
• B (correct): 1/6 minus 1/10 equals 1/15, so a 15 Ω resistor in parallel with 10 Ω gives 6 Ω.
• C: Adding 6 and 10 does not give the missing parallel branch.
• D: Applying product over sum to 6 Ω and 10 Ω combines the target with the existing resistor instead of solving for the missing one.

Answer: A

Correct answer: Each device gets the full line voltage and can work independently of the others. In parallel, every device connects across the same two supply wires, so each receives about 120 V no matter how many others are running. Each device also works on its own, so unplugging one does not affect the others.

• A (correct): Parallel wiring gives every load the same rated voltage, and switching one off does not interrupt the rest.
• B: Dividing the voltage among loads is a series behavior; in parallel each load gets the full voltage.
• C: Each added parallel load lowers the total resistance and raises the total current.
• D: Equal current through every part is a feature of series wiring, and appliances need different currents.

Answer: A

Correct answer: 5 mA. Volts divided by kilohms gives milliamperes. The 4 kΩ branch carries

mA and the 6 kΩ branch carries

mA, so the total is 5 mA, or 0.005 A.

• A (correct): The branches carry 3 mA and 2 mA, which add to 5 mA.
• B: Treating kilohms as ohms makes the current a thousand times too large.
• C: 2.4 is the equivalent resistance in kilohms, not a current.
• D: 12 V divided by 10 kΩ treats the two resistors as if they were in series.

Answer: B

Correct answer: 2 A. The group's reciprocals add to

so the group is 3 Ω. The total is

giving

The series resistor drops 16 V, leaving 12 V across the group, so the 6 Ω branch carries

• A: 4 A is the source current, which splits among the three parallel branches.
• B (correct): The group voltage is 12 V after the 4 Ω drop, and 12 V divided by 6 Ω is 2 A.
• C: 1 A is the current in each 12 Ω branch, not in the 6 Ω branch.
• D: Dividing the full 28 V by 6 Ω ignores the voltage used by the series resistor.

Answer: A

Correct answer: 12 A. The source voltage holds at 12 V, so the 3 Ω and 6 Ω branches still carry 4 A and 2 A. The drifted branch now carries

A instead of 3 A. The total rises from 9 A to

• A (correct): The branches now carry 4 A, 6 A and 2 A, which add to 12 A.
• B: 9 A was the total before the change; the lower-resistance branch now draws more current.
• C: Halving that branch's current assumes current falls with resistance, but at a fixed voltage lower resistance means more current.
• D: 6 A is only the new current in the changed branch; the other two branches still draw their own current.

Answer: B

Correct answer: 12 Ω. Total current is

The 8 Ω branch carries

so the other branch carries

Its resistance is 24 V divided by 2 A, or 12 Ω. Check: 8 parallel with 12 is

and 24 squared divided by 4.8 is 120 W.

• A: 24 squared divided by 120 W gives the equivalent resistance of the whole pair, not the unknown branch.
• B (correct): The total current is 5 A, the 8 Ω branch takes 3 A, and the remaining 2 A at 24 V means 12 Ω.
• C: Subtracting 4.8 Ω from 8 Ω treats the resistances as if they combined in series.
• D: Adding the 3 A branch current to the 5 A total instead of subtracting it gives 8 A and the wrong resistance.

Answer: D

Correct answer: The meter acts as a short across the supply, and a very large current may blow its fuse. An ideal ammeter has nearly zero resistance and belongs in series with the path being measured. Connected across a branch, it lies across both rails and the source, forming a short that can draw a damaging current through the meter.

• A: The low-resistance meter becomes the easiest path, so it carries a very large current rather than none.
• B: To read branch current the meter must be in series with that branch, not across it.
• C: An ammeter measures current through itself; voltage is read with a high-resistance voltmeter across the part.
• D (correct): An ammeter has nearly zero resistance, so placing it across a branch connects the two rails together.

Answer: D

Correct answer: 6 V. Branch B is

so the group is two 6 Ω branches in parallel, or 3 Ω. The total is

so the source current is 3 A and the series resistor drops 3 V, leaving 9 V across each branch. Branch B carries

so the 4 Ω resistor drops 1.5 times 4 = 6 V.

• A: Multiplying the 3 A source current by 4 Ω forgets that the current splits between the two branches.
• B: 9 V is across all of branch B; it is shared between the 2 Ω and 4 Ω resistors.
• C: Applying the full 12 V to branch B ignores the drop across the 1 Ω series resistor.
• D (correct): The group voltage is 9 V, branch B carries 1.5 A, and 1.5 A times 4 Ω is 6 V.

Answer: A

Correct answer: 480 W. Branch currents add in the shared wiring. The microwave draws

A and the lighting draws

for 11 A. The breaker allows 15 A, so 4 A remains, and 120 V times 4 A is 480 W.

• A (correct): The existing loads draw 8 A plus 3 A, or 11 A, leaving 4 A, and 4 A at 120 V is 480 W.
• B: Subtracting only the microwave's 8 A forgets that the lighting also draws current through the breaker.
• C: 1,320 W is the power the existing loads already use, not the remaining room.
• D: 1,800 W is the whole capacity of the circuit; the existing loads already use part of it.

Answer: D

Correct answer: It is four times as large. In series the total is 20 Ω, so

In parallel the total is

Since 80 is four times 20, the parallel connection uses four times the power.

• A: A factor of two comes from counting the resistors, but the equivalent resistance falls from 20 Ω to 5 Ω, a factor of four.
• B: The ratio is reversed; the lower parallel resistance draws more current and more power.
• C: The source voltage is the same, but the equivalent resistances differ, so the total power differs.
• D (correct): Parallel gives 5 Ω and 80 W, while series gives 20 Ω and 20 W, a ratio of four.

Answer: A

Correct answer: 8 A. The equivalent is

so the shared voltage is 14 A times 8/7 Ω, or 16 V. The 2 Ω branch carries

and the others carry 4 A and 2 A, which add to 14 A.

• A (correct): The currents split in the ratio 4 to 2 to 1, so the 2 Ω branch gets 4 of 7 parts of 14 A.
• B: Splitting 14 A equally among three branches ignores the different resistances.
• C: Assigning current in proportion to resistance gives the smallest resistor the least current, which is backward.
• D: The total current divides among all three branches; the 2 Ω branch takes only part of it.

Answer: D

Correct answer: 2 Ω. The series resistor drops 8 A times 3 Ω, or 24 V, leaving 12 V across the group. The 6 Ω branch carries

so R carries

A and

Check: 6 parallel with 2 is 1.5 Ω, plus 3 Ω is 4.5 Ω, and

• A: 1.5 Ω is the equivalent of the whole parallel group, not the unknown branch alone.
• B: 4.5 Ω is the total resistance of the circuit, found from 36 V divided by 8 A.
• C: Using the full 36 V across the parallel group ignores the 24 V dropped by the series resistor.
• D (correct): The group voltage is 12 V, the 6 Ω branch takes 2 A, and the remaining 6 A at 12 V gives 2 Ω.

Answer: A

Correct answer: Total current rises while the current in each original resistor stays the same. Each new branch adds a current path between the same two nodes. The original branches still have the full source voltage, so they carry exactly what they did, and the new branch's current is added on top. More current at the same voltage means a lower equivalent resistance.

• A (correct): The original branches keep the same voltage and current, and the new branch adds its own current to the total.
• B: An ideal source holds its voltage, so adding a branch does not lower the voltage on the others.
• C: With an ideal source, the original branches keep the full voltage, so their currents do not change.
• D: Adding a parallel path always lowers the equivalent resistance.

Answer: B

Correct answer: 20 Ω. For two resistors,

The result is below the smaller resistor, 30 Ω, as it must be.

• A: This adds the resistors as if they were in series.
• B (correct): Product over sum gives 1,800 divided by 90, or 20 Ω.
• C: This averages the two values, which is not how parallel resistors combine.
• D: This adds the reciprocals but forgets the final flip, leaving siemens instead of ohms.

Answer: B

Correct answer: 3 Ω. Use 18 as the common denominator:

Flip the result:

which is below the smallest branch.

• A: This adds the three values as if they were in series.
• B (correct): The reciprocals add to 3/18 + 2/18 + 1/18 = 1/3, so the equivalent is 3 Ω.
• C: This averages the three values, which is not a parallel rule.
• D: This stops at the sum of reciprocals, 1/3, without taking the reciprocal again.

Answer: D

Correct answer: 4 A. For two branches,

For the 12 Ω branch:

Check: the 4 Ω branch carries 12 A, and both branches have 48 V across them.

• A: This is the current in the 4 Ω branch; the larger resistor carries the smaller current.
• B: This splits the current equally, which only happens when the branches are equal.
• C: This gives one branch the full source current, but the current divides among parallel paths.
• D (correct): The 12 Ω branch takes 4/16 of the total, since its share is the other resistance over the sum.

Answer: C

Correct answer: 168 W. Each branch has the full 24 V, so use

W and

Total: 168 W. Check: currents are 4 A and 3 A, and

• A: This adds the resistors to 14 Ω as if they were in series.
• B: This is the power in the 6 Ω branch only.
• C (correct): The branches dissipate 576/6 = 96 W and 576/8 = 72 W, which add to 168 W.
• D: This is the total current of 7 A written as if it were power.

Answer: A

Correct answer: 4 Ω. With the 4 Ω branch open, only 6 Ω and 12 Ω remain:

Fewer paths means a higher equivalent resistance, so 2 Ω rising to 4 Ω makes sense.

• A (correct): The remaining 6 Ω and 12 Ω combine to 72 over 18, or 4 Ω.
• B: This assumes nothing changes, but losing a path raises the equivalent resistance.
• C: This adds the remaining two resistors as if they were in series.
• D: This assumes the equivalent falls, but removing a branch always makes it rise.

Answer: C

Correct answer: The 1 kΩ resistor. Each branch has the battery voltage across it, and

Current is inversely related to resistance, so the 1 kΩ resistor carries the most.

• A: This assumes more resistance means more current, but the largest resistor carries the least.
• B: A middle value carries neither the most nor the least current.
• C (correct): All branches share the same voltage, so the smallest resistance passes the largest current.
• D: This is the second smallest value, so it carries the second largest current.

Answer: D

Correct answer: 90 W. Reduce the parallel pair:

Total resistance is

The series resistor drops 20 V, leaving 30 V across the pair. Then

• A: This puts the full 50 V across the 10 Ω resistor, ignoring the drop in the series resistor.
• B: This is the power in the 15 Ω resistor, which has the same 30 V but more resistance.
• C: This is the power in the 4 Ω series resistor, which carries the full 5 A.
• D (correct): The parallel pair has 30 V across it, so the 10 Ω resistor dissipates 900 divided by 10, or 90 W.

Answer: D

Correct answer: 1 kΩ. Volts over kilohms gives milliamperes: the 1.5 kΩ branch draws

mA.

The other branch carries

mA, so its resistance is

kΩ.

• A: This is 9 V over the total 15 mA, which is the equivalent of both branches together.
• B: This assumes both branches are equal, which would split the current 7.5 mA each.
• C: This has the right digits but drops the kilo prefix when dividing by milliamperes.
• D (correct): The known branch draws 6 mA, leaving 9 mA for the other, and 9 V over 9 mA is 1 kΩ.

Answer: A

Correct answer: 110 Ω. For n equal resistors in parallel,

Here

• A (correct): Two equal resistors in parallel combine to half of one of them.
• B: This adds the two values as if they were in series.
• C: This assumes the second resistor changes nothing, but every added branch lowers the total.
• D: This divides by 4 instead of 2, as if there were four equal branches.

Answer: C

Correct answer: 10 W. Both branches share one voltage, so

makes power inversely proportional to resistance. The 2 Ω branch takes 4 times the power of the 8 Ω branch.

Split 50 W into 5 equal parts of 10 W: the 2 Ω resistor gets 40 W and the 8 Ω resistor gets 10 W.

• A: This is the share of the 2 Ω resistor; in parallel the larger resistor dissipates less.
• B: This splits the power equally, which only happens with equal resistors.
• C (correct): With equal voltage, power is inversely proportional to resistance, so the 8 Ω resistor gets 1 part in 5 of 50 W.
• D: This multiplies 50 W by 2/8 instead of by the 8 Ω branch’s actual share of 1/5.