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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: 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: 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 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: 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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