Correct answer: A — is always less than the smallest resistance

In a parallel circuit, each additional branch provides an extra current path, reducing the total opposition to current flow. The combined (equivalent) resistance is always lower than the smallest individual branch resistance, regardless of how many branches there are or what voltage is applied.

The equivalent resistance for two resistors in parallel is:

\[ R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} \]

For example, with R₁ = 10 Ω and R₂ = 20 Ω:

\[ R_{eq} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} \approx 6.67\ \Omega \]

This result (6.67 Ω) is less than the smallest branch resistance (10 Ω), confirming the rule.

• B. depends upon the voltage drop across each branch — In a parallel circuit all branches share the same voltage; the total resistance is determined by the branch resistances alone, not the applied voltage.
• C. could be equal to the resistance of one branch — The equivalent resistance is always strictly less than the smallest branch; it can never equal any individual branch resistance (unless a branch has infinite resistance, i.e. is open-circuited and effectively not present).
• D. depends upon the applied voltage — Resistance is a property of the components, not the applied voltage. Changing the supply voltage changes current but not resistance (for purely resistive, linear components).

Therefore, the total resistance of a parallel circuit is always less than the smallest individual branch resistance, because parallel paths always increase the total current for a given voltage.

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Correct answer: 80 milliampere

Two equal resistors in parallel have a total resistance of:

\[ R_{\text{total}} = \frac{R}{2} = \frac{1000}{2} = 500\ \Omega \]

Using Ohm’s Law:

\[ I = \frac{V}{R} \]

Given:

• \(V = 40\ \mathrm{V}\)
• \(R_{\text{total}} = 500\ \Omega\)

Substituting:

\[ I = \frac{40}{500} = 0.08\ \mathrm{A} = 80\ \mathrm{mA} \]

• 40 A and 80 A are far too large for this resistance.
• 40 mA would result if the resistors were in series.

Therefore, the total battery current is 80 milliampere.

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Correct answer: sum of the currents through all the parallel branches

In a parallel circuit, current divides among the branches.

The total current supplied by the source is:

\[ I_{\text{total}} = I_1 + I_2 + I_3 + \cdots \]

This follows Kirchhoff’s Current Law.

• It is not equal to just one branch current.
• The other expressions do not correctly describe total current in parallel circuits.

Therefore, the total current is the sum of the currents through all the parallel branches.

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Correct answer: C — series with a resistor

A 3 V bulb connected directly to a 9 V supply would receive three times its rated voltage and burn out immediately. To protect the bulb, a resistor is placed in series with it. The resistor drops the excess voltage (9 V − 3 V = 6 V), leaving only 3 V across the bulb. The same current flows through both components because they are in series.

\[ R = \frac{V_{\text{drop}}}{I} = \frac{6}{I} \]

For example, if the bulb draws 0.1 A, the required series resistor is:

\[ R = \frac{6}{0.1} = 60\ \Omega \]

• Series with the supply is not meaningful on its own — every component must connect to the supply somehow; this option does not describe a voltage-limiting arrangement.
• Parallel with the supply means connecting the bulb directly across the 9 V terminals, which applies the full 9 V to the 3 V bulb and will destroy it.
• Parallel with a resistor places the resistor across the bulb, not in the current path between supply and bulb, so it does not drop the excess 6 V; the full supply voltage still appears across the bulb.

Therefore, connecting the 3 V bulb in series with an appropriately chosen resistor is the correct way to operate it safely from a 9 V supply.

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Correct answer: C — 20

The total current available from the supply must not be exceeded. To find how many lamps can be operated, divide the supply current by the current drawn by each lamp.

\[ N = \frac{I_{\text{supply}}}{I_{\text{lamp}}} \]

Given:

• Supply current: 5 A
• Current per lamp: 250 mA = 0.25 A

\[ N = \frac{5}{0.25} = 20\ \text{lamps} \]

• A. 50 — Incorrect; this would require 50 × 0.25 = 12.5 A, well beyond the 5 A supply.
• B. 30 — Incorrect; 30 × 0.25 = 7.5 A, which exceeds the available supply current.
• D. 5 — Incorrect; this confuses the number of lamps with the supply current value in amps, giving only 5 × 0.25 = 1.25 A used.

Therefore, a 5 A supply can operate exactly 20 lamps each drawing 250 mA before the supply limit is reached.

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Correct answer: A — 12 V

When components are connected in series, the supply voltage is shared across all of them. Each bulb requires its full rated voltage (2 V) to operate normally, so the total supply voltage must equal the sum of all individual voltages.

\[ V_{\text{total}} = V_1 + V_2 + \cdots + V_n \]

With six identical 2 V bulbs:

\[ V_{\text{total}} = 6 \times 2\ \mathrm{V} = 12\ \mathrm{V} \]

• B — 1.2 V: This would be far too low; no bulb would receive even a fraction of its rated voltage.
• C — 6 V: This would only supply 1 V per bulb, so all bulbs would be under-driven and dim.
• D — 2 V: This is the rating of a single bulb only; in a series circuit the voltage divides across all six.

Therefore, a 12 V supply is required to make six series-connected 2 V bulbs light normally.

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Correct answer: 240 / 12

When bulbs are connected in series, the supply voltage is divided equally across them.

To determine how many 12 V bulbs can be used on a 240 V supply:

\[ \text{number of bulbs} = \frac{240}{12} = 20 \]

• Multiplying or adding does not apply.
• Subtracting is not relevant.

Therefore, the correct expression is 240 / 12.

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Correct answer: 30 volt

Three 10,000 \(\Omega\) resistors in series give a total resistance of:

\[ R_{\text{total}} = 10{,}000 + 10{,}000 + 10{,}000 = 30{,}000\ \Omega \]

The same current flows through each resistor in a series circuit.

Using Ohm’s Law:

\[ I = \frac{V}{R} = \frac{90}{30{,}000} = 0.003\ \mathrm{A} \]

Voltage drop across one resistor:

\[ V = IR = 0.003 \times 10{,}000 = 30\ \mathrm{V} \]

• 60 V or 90 V would imply unequal division of voltage.
• 15.8 V does not match the circuit conditions.

Therefore, the voltage drop across one resistor is 30 volt.

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Correct answer: C — 30 ohm

When resistors are connected in parallel, the total (equivalent) resistance is always less than the smallest individual resistor. The standard formula for two resistors in parallel is the product-over-sum rule:

\[ R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2} \]

Substituting the given values:

\[ R_{\text{total}} = \frac{75 \times 50}{75 + 50} = \frac{3750}{125} = 30\ \Omega \]

• 10 ohm — far too low; this would require much smaller resistors or more parallel branches.
• 70 ohm — close to R1 alone; parallel combination always reduces resistance below the smaller resistor (50 Ω here).
• 40 ohm — a plausible-looking distractor but does not result from correct application of the product-over-sum formula.

Therefore, two resistors of 75 Ω and 50 Ω in parallel give a total resistance of 30 Ω.

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Correct answer: A — internal resistance

Every real battery has a small resistance inside it, called internal resistance. When current flows, this internal resistance causes a voltage drop within the cell itself, so the terminal voltage (the voltage available to the external circuit) falls below the open-circuit (no-load) value.

\[ V_{\text{terminal}} = E - I \cdot r \]

Where:

• \(E\) = open-circuit EMF (1.5 V)
• \(I\) = load current
• \(r\) = internal resistance of the cell

The larger the current drawn, the greater the internal voltage drop \(I \cdot r\), and the lower the terminal voltage. Here the drop is 1.5 − 1.2 = 0.3 V, so if you knew the current you could calculate the internal resistance directly (\(r = 0.3 / I\)).

• B. Voltage capacity — not a standard electrical property of a cell; it does not describe the mechanism causing the voltage drop under load.
• C. Electrolyte becoming dry — a dry cell's electrolyte is a paste and does not "dry out" during normal operation; this is not what causes an immediate voltage drop under load.
• D. Current capacity — this refers to the maximum current a cell can deliver (related to its amp-hour rating), not to the voltage drop mechanism.

Therefore, the terminal voltage of a battery drops under heavy load because current flowing through the cell's own internal resistance produces an internal voltage drop, reducing the voltage available to the external circuit.

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Correct answer: A — 5 ohm

When resistors are connected in parallel, the total resistance is always less than the smallest individual resistor. The reciprocal formula is used to find the combined resistance.

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Substituting the values:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{30} = \frac{5}{30} + \frac{1}{30} = \frac{6}{30} = \frac{1}{5} \]

\[ R_{\text{total}} = 5\ \Omega \]

• 8 ohm — this is simply 6 + 2, with no physical basis; it is not the result of any standard series or parallel formula applied to these values.
• 24 ohm — this would result from subtracting 6 from 30, which has no relevance to resistor combination rules.
• 35 ohm — this is 6 + 30, which would be the total for a series connection, not a parallel one.

Therefore, two resistors of 6 Ω and 30 Ω connected in parallel give a total resistance of 5 Ω.

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Correct answer: B — greater than the resistance of any one resistor

When resistors are connected in series, they are joined end-to-end so that the same current flows through each one. Every resistor adds its own opposition to current flow, so the total resistance is the sum of all individual resistances. The result is always larger than any single resistor in the string.

\[ R_{\text{total}} = R_1 + R_2 + R_3 + \cdots \]

Worked example: Three resistors of 10 Ω, 22 Ω, and 47 Ω in series give:

\[ R_{\text{total}} = 10 + 22 + 47 = 79\ \Omega \]

This is greater than any individual resistor (47 Ω being the largest).

• A — less than the resistance of any one resistor: This describes parallel resistors, not series. Parallel connections provide multiple current paths, reducing total resistance.
• C — equal to the highest resistance present: The total is always the sum of all resistors, which exceeds the highest value alone (unless only one resistor is present).
• D — equal to the lowest resistance present: Also incorrect; again, this is not how series addition works.

Therefore, the total resistance of series-connected resistors is always greater than any single resistor because each resistance adds directly to the total.

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Correct answer: D — 50 ohms

When resistors are connected in series, the total resistance is simply the sum of all individual resistances. Each resistor adds its full value to the circuit because the same current must flow through every resistor in turn.

\[ R_{\text{total}} = R_1 + R_2 + R_3 + R_4 + R_5 \]

With five 10 Ω resistors:

\[ R_{\text{total}} = 10 + 10 + 10 + 10 + 10 = 50\ \Omega \]

• A — 1 ohm: This would result from five 5 Ω resistors in parallel, not series.
• B — 5 ohms: This is the result of dividing one resistor's value by the number of resistors — a common confusion with the parallel formula.
• C — 10 ohms: This equals the value of just one resistor; adding more resistors in series always increases the total resistance.

Therefore, five 10 Ω resistors in series give a total resistance of 50 Ω.

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Correct answer: C — 4280 ohm

When resistors are connected in series, the total resistance is simply the sum of all individual resistances. Each resistor adds directly to the total because the same current must flow through every component in turn.

\[ R_{\text{total}} = R_1 + R_2 + R_3 + R_4 \]

Substituting the given values:

\[ R_{\text{total}} = 10 + 270 + 3900 + 100 = 4280\ \Omega \]

• A — 9 ohm: This is far too low; it is approximately what you would get if the resistors were combined in a parallel arrangement (and even then it would not be correct for these values).
• B — 3900 ohm: This is only the largest single resistor in the group, ignoring the other three.
• D — 10 ohm: This is only the smallest single resistor in the group, ignoring the other three.

Therefore, four resistors of 10, 270, 3900, and 100 ohm connected in series give a total resistance of 4280 ohm.

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Correct answer: five 24 ohm

In a series circuit, resistances add directly:

\[ R_{\text{total}} = R_1 + R_2 + \cdots \]

For five 24 \(\Omega\) resistors:

\[ R_{\text{total}} = 5 \times 24 = 120\ \Omega \]

• Six 22 \(\Omega\) = 132 \(\Omega\)
• Two 62 \(\Omega\) = 124 \(\Omega\)
• Five 100 \(\Omega\) = 500 \(\Omega\)

Therefore, the correct combination is five 24 ohm.

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Correct answer: 2.3 megohm

Convert both resistances to the same units:

\[ 2.2\ \mathrm{M}\Omega = 2{,}200{,}000\ \Omega \]

\[ 100\ \mathrm{k}\Omega = 100{,}000\ \Omega \]

In series:

\[ R_{\text{total}} = 2{,}200{,}000 + 100{,}000 = 2{,}300{,}000\ \Omega = 2.3\ \mathrm{M}\Omega \]

• 2.1 M\(\Omega\) and 2.11 M\(\Omega\) are too low.
• 2.21 M\(\Omega\) does not correctly add the values.

Therefore, the total resistance is 2.3 megohm.

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Correct answer: R/10

For resistors in parallel:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R} + \frac{1}{R} + \cdots \text{(10 times)} \]

So:

\[ \frac{1}{R_{\text{total}}} = \frac{10}{R} \quad \Rightarrow \quad R_{\text{total}} = \frac{R}{10} \]

• R would apply to a single resistor.
• 10R applies to series connection.
• 10/R is not dimensionally correct.

Therefore, the total resistance is R/10.

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Correct answer: 17 ohm

For resistors in parallel:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \]

Since all resistors are equal (68 \(\Omega\)):

\[ R_{\text{total}} = \frac{68}{4} = 17\ \Omega \]

• 12 \(\Omega\) is too low.
• 34 \(\Omega\) would be two in parallel.
• 272 \(\Omega\) would be four in series.

Therefore, the total resistance is 17 ohm.

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Correct answer: A — less than 10 ohm

When resistors are connected in parallel, the total (equivalent) resistance is always less than the smallest individual resistor. Here the smallest resistor is 10 Ω, so the total must be less than 10 Ω.

The governing formula for parallel resistance is:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \]

Substituting the values:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{68} + \frac{1}{47000} + \frac{1}{560} + \frac{1}{10} \]

\[ \frac{1}{R_{\text{total}}} \approx 0.01471 + 0.00002 + 0.00179 + 0.10000 = 0.11652 \]

\[ R_{\text{total}} \approx 8.6\ \Omega \]

• B. between 68 and 560 ohm — Incorrect; parallel combinations always reduce total resistance below the smallest resistor, not somewhere in the middle of the range.
• C. between 560 and 47 kilohm — Incorrect; same reasoning; adding parallel paths only decreases total resistance.
• D. greater than 47 kilohm — Incorrect; this would only be possible for series resistors, not parallel.

Therefore, the total resistance of these four resistors in parallel is approximately 8.6 Ω, which is less than the smallest resistor (10 Ω).

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Correct answer: C — three 47 ohm resistors in series

Resistors in series simply add together. Three 47 Ω resistors placed in series give a total resistance very close to 150 Ω, making this the best substitution.

\[ R_{\text{total}} = R_1 + R_2 + R_3 = 47 + 47 + 47 = 141\ \Omega \]

141 Ω is the closest approximation to 150 Ω among the options given.

For resistors in parallel, the combined resistance is always less than the smallest individual resistor, so parallel combinations of 33 Ω or 47 Ω resistors will produce values well below 150 Ω:

• A. Four 47 Ω resistors in parallel — gives 47/4 = 11.75 Ω, far too low.
• B. Five 33 Ω resistors in parallel — gives 33/5 = 6.6 Ω, far too low.
• D. Five 33 Ω resistors in series — gives 5 × 33 = 165 Ω, closer but still further from 150 Ω than option C (141 Ω vs 165 Ω; errors of −9 Ω vs +15 Ω).

Therefore, three 47 Ω resistors wired in series, totalling 141 Ω, most nearly replaces a single 150 Ω resistor.

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Correct answer: C — 60 ohm

When two equal resistors are connected in parallel, the combined resistance is exactly half the value of either individual resistor. So two 120 Ω resistors in parallel produce 60 Ω, which is the value of the faulty resistor being replaced.

\[ R_{\text{parallel}} = \frac{R_1 \times R_2}{R_1 + R_2} \]

\[ R_{\text{parallel}} = \frac{120 \times 120}{120 + 120} = \frac{14400}{240} = 60\ \Omega \]

• A — 15 Ω: This would require four 60 Ω resistors in parallel, or some other combination; two 120 Ω resistors cannot produce 15 Ω.
• B — 30 Ω: This would require two 60 Ω resistors in parallel, not two 120 Ω resistors.
• D — 120 Ω: Two 120 Ω resistors in parallel give half of 120 Ω, not 120 Ω itself; a single 120 Ω resistor would be needed to replace a 120 Ω component.

Therefore, two 120 Ω resistors connected in parallel produce a combined resistance of 60 Ω, which is the value of the original faulty resistor.

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Correct answer: A has half the resistance of B

For resistors in parallel, the voltage across each resistor is the same.

Using Ohm’s Law:

\[ I = \frac{V}{R} \]

Since both resistors have the same voltage across them, the current through each is inversely proportional to its resistance.

If resistor A carries twice the current of resistor B:

\[ I_A = 2 I_B \]

then:

\[ \frac{V}{R_A} = 2 \times \frac{V}{R_B} \Rightarrow R_A = \frac{R_B}{2} \]

• The voltage across both resistors is equal.
• B does not have half the resistance of A.

Therefore, resistor A has half the resistance of resistor B.

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Correct answer: 200 ohm by arranging them in parallel

The smallest possible resistance is obtained by connecting all resistors in parallel.

For \(n\) equal resistors in parallel:

\[ R_{\text{total}} = \frac{R}{n} \]

Given:

• \(R = 1000\ \Omega\)
• \(n = 5\)

Substituting:

\[ R_{\text{total}} = \frac{1000}{5} = 200\ \Omega \]

• Series connection would increase the total resistance.
• 50 \(\Omega\) is not achievable with five 1 k\(\Omega\) resistors in either series or parallel.

Therefore, the smallest resistance is 200 ohm by arranging them in parallel.

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Correct answer: a combination of two resistors in parallel, then placed in series with another resistor

Two 28 \(\Omega\) resistors in parallel give:

\[ R_{\text{parallel}} = \frac{28 \times 28}{28 + 28} = \frac{784}{56} = 14\ \Omega \]

Placing this in series with another 28 \(\Omega\) resistor:

\[ R_{\text{total}} = 14 + 28 = 42\ \Omega \]

• Three in series would give 84 \(\Omega\).
• Three in parallel would give about 9.3 \(\Omega\).
• Two parallel pairs in series would give 28 \(\Omega\).

Therefore, the correct combination gives 42 ohm.

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Correct answer: A — 60 ohms

When two equal resistors are connected in parallel, the combined resistance is half the value of one resistor. That parallel combination is then added directly to the series resistor.

Step 1 — Parallel combination of the two 100 Ω resistors:

\[ R_{\text{parallel}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{100 \times 100}{100 + 100} = \frac{10000}{200} = 50\ \Omega \]

Step 2 — Add the series 10 Ω resistor:

\[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{series}} = 50 + 10 = 60\ \Omega \]

• B — 180 ohms: Incorrect; this would result from adding all three resistors in series (100 + 100 − 10 = 190, or similar arithmetic errors), not accounting for the parallel arrangement.
• C — 190 ohms: Incorrect; this is simply 100 + 100 − 10, ignoring the parallel rule entirely.
• D — 210 ohms: Incorrect; this appears to be 100 + 100 + 10, treating all three resistors as being in series.

Therefore, the total resistance of the two 100 Ω resistors in parallel (giving 50 Ω) wired in series with a 10 Ω resistor is 60 ohms.

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Correct answer: B — 1 ampere

When resistors are connected in series, their resistances add directly to give the total resistance. Ohm's Law then gives the current drawn from the supply.

\[ R_{\text{total}} = R_1 + R_2 = 5 + 10 = 15\ \Omega \]

\[ I = \frac{V}{R_{\text{total}}} = \frac{15}{15} = 1\ \text{A} \]

• A (0.5 A): This would require a total resistance of 30 Ω — incorrect for these two resistors in series.
• C (2 A): This would require a total resistance of 7.5 Ω, which would only be possible if the resistors were combined differently (e.g. partially in parallel).
• D (15 A): This would imply a total resistance of only 1 Ω, ignoring both resistors entirely.

Therefore, with 5 Ω and 10 Ω in series giving 15 Ω total, and 15 V applied, Ohm's Law gives exactly 1 ampere.

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Correct answer: 2 amperes

Three equal resistors in parallel:

\[ R_{\text{total}} = \frac{R}{n} = \frac{12}{3} = 4\ \Omega \]

Using Ohm’s Law:

\[ I = \frac{V}{R} = \frac{8}{4} = 2\ \mathrm{A} \]

• 1 A would require higher resistance.
• 3 A and 4.5 A are too high for this setup.

Therefore, the total current is 2 amperes.

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Correct answer: C — 3.3 volt

When two equal resistors are connected in series, the total resistance is their sum. The voltage across each individual resistor is found using Ohm's Law: V = IR, where I is the current through the circuit and R is the resistance of that one resistor.

\[ V = I \times R \]

Given:

• R (one resistor) = 33 Ω
• I = 100 mA = 0.1 A

\[ V = 0.1 \times 33 = 3.3\ \mathrm{V} \]

• A. 66 volt — This would be the total voltage across both resistors in series (0.1 × 66 = 6.6 V), not even that; 66 V would require 1 A through 66 Ω — neither value applies here.
• B. 33 volt — This confuses the resistance value (33 Ω) with the voltage; it ignores the actual current of 100 mA.
• D. 1 volt — This has no basis in the given values; it may arise from incorrectly dividing or rounding.

Therefore, applying Ohm's Law to a single 33 Ω resistor with 100 mA flowing through it gives a voltage drop of 3.3 V.

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Correct answer: C — six 300 ohm resistors in parallel

When identical resistors are connected in parallel, the combined resistance equals the single resistor value divided by the number of resistors. To achieve a 50 Ω dummy load from 300 Ω resistors, divide 300 by the number of resistors and solve for when the result equals 50 Ω.

\[ R_{\text{parallel}} = \frac{R}{n} \]

\[ 50 = \frac{300}{n} \Rightarrow n = \frac{300}{50} = 6 \]

Six 300 Ω resistors in parallel gives exactly 50 Ω.

• Four 300 Ω resistors in parallel gives 300/4 = 75 Ω — close to common coaxial impedance but not 50 Ω.
• Five 300 Ω resistors in parallel gives 300/5 = 60 Ω — still too high.
• Seven 300 Ω resistors in parallel gives 300/7 ≈ 42.9 Ω — too low.

Therefore, six 300 Ω resistors wired in parallel is the correct combination to fabricate a 50 Ω dummy load for testing a transmitter safely.

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Correct answer: 1500 ohm to 1000 ohm

Initially, three 500 \(\Omega\) resistors in series:

\[ R_{\text{initial}} = 500 + 500 + 500 = 1500\ \Omega \]

If the centre resistor is short-circuited, it is effectively bypassed (zero resistance), leaving:

\[ R_{\text{new}} = 500 + 500 = 1000\ \Omega \]

• The total resistance decreases because one resistor is removed from the circuit.
• Other options do not match the correct series calculation.

Therefore, the network changes from 1500 ohm to 1000 ohm.

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