1. Physics Circuits: Current / Voltage / Resistance
FACULTY OF EDUCATION
Department of
Curriculum and Pedagogy
Physics Circuits: Current / Voltage / Resistance
Category: Secondary – Physics – Circuits
Tags: voltage, current, resistance, parallel circuits, series circuits, power
Excerpt: The behavior of voltage, current and resistance will be analyzed in both series and parallel circuits.
Science and Mathematics Education Research Group
Supported by UBC Teaching and Learning Enhancement Fund

2. Parallel and Series Circuits+ – + –

3. Series and Parallel Circuits I
Which of the two circuits below is a series circuit and which is a parallel circuit? + – I.
I – Series, II – Parallel
I – Parallel, II – Series
Both are series circuits
Both are parallel circuits
Depends on the battery + –
II.

4. Solution I. Series circuit II. Parallel circuit + + – – Answer: A
Justification: The series circuit has no junctions (branches) between consecutive circuit elements while there are in the parallel circuit. The exact properties of series and parallel circuits that define series and parallel circuits will be explored in the following questions.
I. Series circuit
II. Parallel circuit
Junctions
Which of the following circuits are parallel….. + + – –

5. Series and Parallel Circuits II
How many of the circuits below are parallel circuits? + – I.
II. + –
III.
IV. + + – –
A. Zero
C. Two
B. One
D. Three
E. Four

6. Solution
Answer: E
Justification: If the wires in the circuits have negligible resistance, then all the circuits will be identical. The wires in circuit I can be bent to closely resemble the other circuits without reconnecting wires.

7. Series Circuits III
Assume that 100 electrons flow out of the negative terminal of the battery into Bulb B over a period of time. The bulb dissipates energy, creating heat and light in the process. How many electrons flow out of Bulb B? ΔV
Bulb A
Bulb B I
Electron flow
More than 100 electrons
Exactly 100 electrons
Fewer than 100 electrons
Exactly 0 electrons
Depends on the resistance of the bulbs

8. Solution
Answer: B
Justification: The same number of electrons that enter the light bulb must also leave the light bulb. Electrons are not “used up” in a light bulb. The electrons have nowhere else to go.
Student tip: Imagine a circuit carrying electrons is like a hose carrying water. The rate that water enters a device must be equal to the rate that water leaves a device.

9. Series Circuits IV Bulb A Iinitial + IA – IB Bulb B
A current Iinitial flows from the positive terminal of the battery into Bulb A. The current IA leaves Bulb A and enters Bulb B. A current IB leaves Bulb B and returns to the battery. Which of the following correctly describes the relationship between IInitial, IA and IB?
Iinitial + – IA IB
Bulb A
Bulb B
Iinitial > IA > IB
Iinitial = IA = IB
Iinitial < IA < IB
Iinitial = IA + IB
Iinitial = IA – IB

10. Solution
Answer: B
Justification: Current is defined as the amount of charge that passes through a point in a wire per unit of time.
From the previous question, we know that no charge is lost when electrons move in a circuit. Therefore, the current must be the same at all points along the single path in a series circuit.

11. Series Circuits V
The battery in the circuit has a voltage of 9 V. If the electric potential at point A is VA = 9 V, what is the electric potential at point B? (The light bulbs do not necessarily have the same resistance.)
ΔV = 9 V
VB = ?
VA = 9 V
Exactly 9 V
Between 9 V and 4.5 V
Exactly 4.5 V
Between 4.5 V and 0 V
Exactly 0 V

12. Solution VA = 9 V ΔVB to A= VA – VB = 9 V VB = 0 V Answer: E
Justification: The potential at the positive terminal of the battery is equal to the potential at point A, while the potential at the negative terminal is equal to the potential at point B.
If the potential difference across the terminals of the battery is 9 V, the potential at B must be 0 V.
ΔVB to A= VA – VB = 9 V
VB = 0 V
VA = 9 V

13. Series Circuits VI
The battery in the circuit has a voltage of 9 V. The resistance of Bulb A is 2R while the resistance of Bulb B is R. If the electric potential at point A is VA = 9 V, what is the electric potential at point B?
Bulb A (2R)
Bulb B (R)
VB = 9 V
VB = 6 V
VB = 4.5 V
VB = 3 V
VB = 0 V
VA = 9 V
VB = ?
ΔV = 9 V

14. Solution 9 V 3 V 0 V ΔV = 9 V Answer: D
Justification: From the previous question, we know the potential drop across both bulbs must total –9 V (to go from 9 V to 0 V). Since ΔV = IR and Bulb A has twice the resistance of Bulb B, the potential drop across Bulb A must be twice as large as Bulb B. This is only possible when ΔVA = –6 V and ΔVB = –3 V
In a series circuit, current is the same across resistors, but the voltage may be different
9 V R
3 V R
0 V
Notice that both bulbs taken together are equivalent to a 3R bulb.
ΔVA = –6 V
ΔVB = –3 V
ΔV = 9 V

15. Series Circuits VII
Two identical light bulbs are connected in a parallel circuit as shown in the diagram. How will the bulbs’ brightness compare with one another? ΔV
Bulb A
Bulb B R
Bulb A is brighter than Bulb B
Bulb B is brighter than Bulb A
The bulbs are equally bright
Bulb A is lit while Bulb B is not lit
Bulb B is lit while Bulb A is not lit

16. Solution
Answer: C
Justification: The bulbs are identical so the resistance of each bulb must be the same. We know from question III that the current is the same across both light bulbs. Since ΔV = IR, the potential difference across each light bulb is the same if the current and resistance is the same.
Since the voltage, current, and resistance across the two bulbs are the same, they use the same amount of power and are therefore equally bright.

17. Parallel Circuits VIII
Current Iinitial leaves the positive terminal of the battery. The current then splits into IA and IB as shown. Current Ifinal enters the negative terminal of the battery. Which of the following correctly describes the relationship between Iinitial, IA, IB and Ifinal? (The light bulbs do not necessarily have the same resistance.)
Bulb A
Bulb B + –
Iinitial IA IB
Ifinal
Iinitial > IA > IB > Ifinal
Iinitial = 2IA = 2IB = Ifinal
Iinitial = IA + IB = Ifinal
Iinitial = IA – IB = Ifinal
Depends on the resistance of the bulbs

18. Solution
Answer: C
Justification: The current in a circuit must be conserved. The current Iinitial splits into IA and IB. Therefore the sum of IA and IB must equal Iinitial. Likewise, currents IA and IB converge to form Ifinal. Therefore:
Iinitial = IA + IB = Ifinal
Bulb A
Bulb B + –
Iinitial IA IB
Ifinal
Only when the bulbs have same resistance will the current split evenly in half.

19. Parallel Circuits IX
The voltage across the battery is 9 V. If the electric potential at point A is VA = 9 V, what is the electric potential at point B? (The bulbs do not necessarily have the same resistance.)
Exactly 9 V
Between 9 V and 4.5 V
Exactly 4.5 V
Between 4.5 V and 0 V
Exactly 0 V
VA = 9 V ΔV
VB = ?

20. Solution VA = 9 V ΔVB to A = VA – VB = 9 V VB = 0 V Answer: E
Justification: There is no resistance across along the red wires so there is no drop in potential. Therefore all points along the red wire must have a potential of 9 V. Similarly, all points along the blue wires must also have the same potential.
In order for there to be a potential difference of 9 V across the battery, point B must have a potential of 0 V. Notice that the potential difference across every bulb is 9 V.
ΔVB to A = VA – VB = 9 V
VA = 9 V
VB = 0 V

21. Parallel Circuits X
Assume that the potential at point A is 9 V while the potential at point B is 0 V. Bulb A has a resistance of 2R while Bulb B has a resistance of R.
What is the potential difference across Bulb A (ΔVA) and Bulb B (ΔVB)?
VB = 0 V ΔV
VA = 9 V
Bulb A (2R)
Bulb B (R)
ΔVA = –9 V, ΔVB = –9 V
ΔVA = –9 V, ΔVB = –4.5 V
ΔVA = –9 V, ΔVB = 0 V
ΔVA = –4.5 V, ΔVB = –4.5 V
ΔVA = –4.5V, ΔVB = 0 V

22. Solution VA = 9 V ΔVA to B VB = 0 V Answer: A
Justification: From the previous question, we know the potential along the wires are as shown.
The potential at the top end of each bulb is VA = 9 V. The potential at the bottom end of each bulb is VB = 0 V. The potential difference across each bulb is the difference between VA and VB, or ΔVA to B = VB – VA = –9 V. The negative sign tells us that there is a drop in potential across the bulb when following the direction of the current.
VA = 9 V
VB = 0 V
ΔVA to B

23. Parallel Circuits XI
Two identical light bulbs are connected in a series circuit as shown in the diagram. How will the bulbs’ brightness compare with one another?
Bulb A is brighter than Bulb B
Bulb B is brighter than Bulb A
The bulbs are equally bright
Bulb A is lit but Bulb B is not
Bulb B is lit but Bulb A is not ΔV
Bulb A
Bulb B R

24. Solution IT IA IB ΔV R Answer: C
Justification: We know from the previous question that the potential difference across both bulbs is ΔV.
Since the potential difference is the same across both bulbs, the current IA and IB depends only on the resistance through each path. Since the bulbs have the same resistance, the current through the bulbs is the same. ΔV R IT IA IB
Since there is the voltage, current, and resistance across each bulb, they must dissipate the same amount of energy and are therefore equally bright.

25. Series and Parallel Circuits XII
Two identical light bulbs are connected first in a series circuit and then in a parallel circuit with the same battery. In which circuit will the bulbs be brighter? ΔV R ΔV R
The bulbs will be brighter in the series circuit
The bulbs will be brighter in the parallel circuit
The bulbs will be equally bright in both circuits

26. Series and Parallel Circuits XIII
Two identical light bulbs are connected first in a series circuit and then in a parallel circuit with the same battery. Which of the follow equations best describes the relative amount of power dissipated in each circuit?
Series: ΔV R
Parallel: ΔV R

27. Solution Answer: XII) B, XIII) A Justification:
We know from previous questions the potential difference across the bulbs in parallel is ΔV. The potential difference across the bulbs in series must be 0.5ΔV since the bulbs have the same resistance. Since the power dissipated by a bulb is:
we can determine the parallel circuit will use 4 times the amount of energy.
Notice that we do not need to calculate the current through the bulbs.

28. Alternate Solution Part I
Answer: XII) B, XIII) A
Justification: The current across the 2 bulbs is The potential difference across each bulb is Using any formula for power gives: Since there are 2 bulbs, the total power dissipated in the circuit is: ΔV R

29. Alternate Solution Part II
Answer: IX) B, X) A
Justification: Parallel circuit:
The potential difference across each bulb must be ΔV and the resistance of each bulb is R. The power dissipated in each bulb is The total power dissipated in the circuit is:
Compared to the series circuit: ΔV R R
The bulbs in parallel dissipate 4 times as much power and are therefore brighter.