Below is a set of revision test questions for APPLIED SCIENCE Unit 15
(Electrical Circuits and their Applications), with each question
followed by a brief rationale explaining the answer. You can use these
to self‐assess your understanding of key concepts from the unit.
Revision Test Questions
Question 1 – Electrical Symbols and Definitions
Match each of the following circuit components with its correct symbol:
A. Cell
B. Battery
C. Switch (closed)
D. Fixed resistor
E. Filament lamp
Options:
1. A circle with a plus and minus sign inside
2. A series of small “zigzags”
3. Two or more cells connected together
4. A break in the circuit with a line bridging the gap
5. A curved line (representing a light bulb)
Rationale:
Understanding and recalling the standard symbols is fundamental. For example, a cell is typically
represented as a single cell with a longer positive line and a shorter negative line; a battery is a series of
such cells; a resistor is shown as a zigzag line; a switch in its closed position is shown as a break with a
connecting line; and a filament lamp symbol often resembles a small circle with filament lines.
Question 2 – Ohm’s Law Calculation
A resistor with a resistance of 6.0 Ω is connected across a 12 V power supply. What is the current
flowing through the resistor?
A. 0.5 A B. 1.0 A C. 2.0 A D. 4.0 A
Rationale:
Ohm’s law states that V=IRV = IRV=IR. Rearranging gives I=VRI = \frac{V}{R}I=RV. Substituting in the
numbers:
I=12 V6.0 Ω=2.0 AI = \frac{12\,V}{6.0\,Ω} = 2.0\,AI=6.0Ω12V=2.0A
This shows the correct answer is C. Knowing how to rearrange and apply Ohm’s law is critical for
predicting circuit behavior.
, Question 3 – Kirchhoff’s First Law
At a junction in a circuit, two currents leave the junction, measuring 3 A and 4 A. If the current entering
the junction is IinI_{\text{in}}Iin, what is IinI_{\text{in}}Iin?
A. 1 A B. 3 A C. 7 A D. 12 A
Rationale:
Kirchhoff’s Current Law (KCL) states that the total current entering a junction equals the total current
leaving it. Here, Iin=3 A+4 A=7 AI_{\text{in}} = 3\,A + 4\,A = 7\,AIin=3A+4A=7A. This simple summing
demonstrates conservation of charge in circuits.
Question 4 – Resistors in Series and Parallel
You have three resistors: 100 Ω, 200 Ω, and 300 Ω. Calculate:
a) The total resistance when they are connected in series.
b) The total resistance when they are connected in parallel.
Options:
a) Series:
A. 600 Ω B. 300 Ω C. 200 Ω D. 100 Ω
b) Parallel (approximate value):
A. 50 Ω B. 85 Ω C. 600 Ω D. 1000 Ω
Rationale:
For series: Rtotal(series)=100+200+300=600 ΩR_{\text{total(series)}} = 100 + 200 + 300 =
600\,ΩRtotal(series)=100+200+300=600Ω (Answer a: A).
For parallel:
1Rtotal(parallel)=1100+1200+1300\frac{1}{R_{\text{total(parallel)}}} = \frac{1}{100} + \frac{1}{200} +
\frac{1}{300}Rtotal(parallel)1=1001+2001+3001
1Rtotal(parallel)=0.01+0.005+0.00333≈0.01833\frac{1}{R_{\text{total(parallel)}}} = 0.01 + 0.005 +
0.00333 \approx 0.01833Rtotal(parallel)1=0.01+0.005+0.00333≈0.01833
Thus,
Rtotal(parallel)≈10.01833≈54.6 ΩR_{\text{total(parallel)}} \approx \frac{1}{0.01833} \approx
54.6\,ΩRtotal(parallel)≈0.018331≈54.6Ω
Closest option is approximately 50 Ω (Answer b: A). This reinforces how resistor combinations affect
overall resistance.
Question 5 – Capacitance in Series and Parallel
Three capacitors have values: 10 µF, 20 µF, and 30 µF. Calculate:
a) The total capacitance when connected in parallel.
b) The total capacitance when connected in series.
(Electrical Circuits and their Applications), with each question
followed by a brief rationale explaining the answer. You can use these
to self‐assess your understanding of key concepts from the unit.
Revision Test Questions
Question 1 – Electrical Symbols and Definitions
Match each of the following circuit components with its correct symbol:
A. Cell
B. Battery
C. Switch (closed)
D. Fixed resistor
E. Filament lamp
Options:
1. A circle with a plus and minus sign inside
2. A series of small “zigzags”
3. Two or more cells connected together
4. A break in the circuit with a line bridging the gap
5. A curved line (representing a light bulb)
Rationale:
Understanding and recalling the standard symbols is fundamental. For example, a cell is typically
represented as a single cell with a longer positive line and a shorter negative line; a battery is a series of
such cells; a resistor is shown as a zigzag line; a switch in its closed position is shown as a break with a
connecting line; and a filament lamp symbol often resembles a small circle with filament lines.
Question 2 – Ohm’s Law Calculation
A resistor with a resistance of 6.0 Ω is connected across a 12 V power supply. What is the current
flowing through the resistor?
A. 0.5 A B. 1.0 A C. 2.0 A D. 4.0 A
Rationale:
Ohm’s law states that V=IRV = IRV=IR. Rearranging gives I=VRI = \frac{V}{R}I=RV. Substituting in the
numbers:
I=12 V6.0 Ω=2.0 AI = \frac{12\,V}{6.0\,Ω} = 2.0\,AI=6.0Ω12V=2.0A
This shows the correct answer is C. Knowing how to rearrange and apply Ohm’s law is critical for
predicting circuit behavior.
, Question 3 – Kirchhoff’s First Law
At a junction in a circuit, two currents leave the junction, measuring 3 A and 4 A. If the current entering
the junction is IinI_{\text{in}}Iin, what is IinI_{\text{in}}Iin?
A. 1 A B. 3 A C. 7 A D. 12 A
Rationale:
Kirchhoff’s Current Law (KCL) states that the total current entering a junction equals the total current
leaving it. Here, Iin=3 A+4 A=7 AI_{\text{in}} = 3\,A + 4\,A = 7\,AIin=3A+4A=7A. This simple summing
demonstrates conservation of charge in circuits.
Question 4 – Resistors in Series and Parallel
You have three resistors: 100 Ω, 200 Ω, and 300 Ω. Calculate:
a) The total resistance when they are connected in series.
b) The total resistance when they are connected in parallel.
Options:
a) Series:
A. 600 Ω B. 300 Ω C. 200 Ω D. 100 Ω
b) Parallel (approximate value):
A. 50 Ω B. 85 Ω C. 600 Ω D. 1000 Ω
Rationale:
For series: Rtotal(series)=100+200+300=600 ΩR_{\text{total(series)}} = 100 + 200 + 300 =
600\,ΩRtotal(series)=100+200+300=600Ω (Answer a: A).
For parallel:
1Rtotal(parallel)=1100+1200+1300\frac{1}{R_{\text{total(parallel)}}} = \frac{1}{100} + \frac{1}{200} +
\frac{1}{300}Rtotal(parallel)1=1001+2001+3001
1Rtotal(parallel)=0.01+0.005+0.00333≈0.01833\frac{1}{R_{\text{total(parallel)}}} = 0.01 + 0.005 +
0.00333 \approx 0.01833Rtotal(parallel)1=0.01+0.005+0.00333≈0.01833
Thus,
Rtotal(parallel)≈10.01833≈54.6 ΩR_{\text{total(parallel)}} \approx \frac{1}{0.01833} \approx
54.6\,ΩRtotal(parallel)≈0.018331≈54.6Ω
Closest option is approximately 50 Ω (Answer b: A). This reinforces how resistor combinations affect
overall resistance.
Question 5 – Capacitance in Series and Parallel
Three capacitors have values: 10 µF, 20 µF, and 30 µF. Calculate:
a) The total capacitance when connected in parallel.
b) The total capacitance when connected in series.
