Chapter 1 · Connecting resistors & Kirchhoff's laws: practice questions
Unit 1 — Dynamic Electricity and Electromagnetism · Physics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 85 questions on this lesson, each with an explanation.
Connecting an additional resistor in parallel with an existing combination of resistors causes the equivalent resistance of the whole combination to…
Adecrease
Bbecome infinite
Cremain constant
Dincrease
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Correct answer: (A) decrease
Explanation: An added parallel branch provides an extra path for the current, lowering the equivalent resistance.
Question 2Easy
In a parallel combination of resistors connected across a source, which of the following quantities is the same for every resistor?
Athe electric power dissipated in it
Bthe value of its resistance
Cthe potential difference across it
Dthe electric current through it
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Correct answer: (C) the potential difference across it
Explanation: All branches of a parallel combination are connected between the same two points, so they share the same potential difference.
Question 3Easy
Two identical resistors, each of resistance 12 Ω, are connected in parallel. The equivalent resistance of the combination is equal to…
A12 Ω
B3 Ω
C24 Ω
D6 Ω
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Correct answer: (D) 6 Ω
Explanation: Two equal resistors R in parallel give an equivalent resistance R/2 = 12/2 = 6 Ω.
Question 4Medium
Three equal resistors, each of resistance R, are available. To obtain the smallest possible equivalent resistance, they should be connected…
Atwo in parallel then in series with the third
Ball in series
Call in parallel
Dtwo in series then in parallel with the third
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Correct answer: (C) all in parallel
Explanation: Connecting all three in parallel gives R/3, the smallest equivalent resistance obtainable from the three resistors.
Question 5Medium
Two resistors of resistances R and 3R are connected in series. The ratio between the potential difference across R and that across 3R is equal to…
A1/1
B1/9
C3/1
D1/3
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Correct answer: (D) 1/3
Explanation: In series the current is common and V ∝ R, so VR : V_(3R) = 1 : 3.
Question 6Medium
A 6 Ω resistor and a 3 Ω resistor are connected in parallel across a battery of e.m.f. 12 V (VB) and negligible internal resistance. The total current drawn from the battery is equal to…
A6 A
B4 A
C12 A
D2 A
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Correct answer: (A) 6 A
Explanation: The parallel combination is 2 Ω, so the total current is I = 12/2 = 6 A.
Question 7Medium
What resistance must be connected in parallel with a 12 Ω resistor so that the equivalent resistance of the combination becomes 4 Ω?
A6 Ω
B16 Ω
C3 Ω
D8 Ω
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Correct answer: (A) 6 Ω
Explanation: 1/R = 1/4 − 1/12 = 1/6, so R = 6 Ω.
Question 8Hard
Two batteries are connected in parallel (positive to positive) across a 4 Ω resistor. Battery A has VB = 12 V and battery B has VB = 6 V, and each has an internal resistance of 1 Ω. The current in battery B and its state are…
A4 A, and battery B is discharging
B1.5 A, and battery B is being charged
C2 A, and battery B is discharging
D2 A, and battery B is being charged
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Correct answer: (D) 2 A, and battery B is being charged
Explanation: With V the common terminal p.d., Kirchhoff's first law gives (12 − V)/1 + (6 − V)/1 = V/4, so V = 8 V. The current through B is (6 − 8)/1 = −2 A: 2 A is forced into B against its e.m.f., so it is being charged.
Question 9Hard
A total current of 12 A is fed into a parallel combination of three resistors of resistances 2 Ω, 3 Ω and 6 Ω. The current passing through the 3 Ω resistor is equal to…
A4 A
B12 A
C2 A
D6 A
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Correct answer: (A) 4 A
Explanation: The parallel equivalent is 1 Ω, so the common p.d. = 12 × 1 = 12 V and I(3Ω) = 12/3 = 4 A.
Question 10Hard
A resistor R is connected in parallel with an 8 Ω resistor, and this parallel combination is connected in series with a 6 Ω resistor. If the total equivalent resistance is 10 Ω, the value of R is equal to…
A8 Ω
B16 Ω
C2 Ω
D4 Ω
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Correct answer: (A) 8 Ω
Explanation: The parallel part = 10 − 6 = 4 Ω, so 1/R = 1/4 − 1/8 = 1/8, giving R = 8 Ω.