WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
D.C. Circuits
Working out currents and potential differences in resistor networks
- 2 things to know
- AS Unit 2
Overview
Recall questions on Working out currents and potential differences in resistor networks.
What you need to know
Answer each one before you reveal it.
What happens to the total resistance of a network when a further resistor is connected in parallel with part of it?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Describe the general method for finding the current in every part of a network of resistors connected to a supply.
Model answer
The network is first reduced step by step to a single equivalent resistance by combining series and parallel groups, the current drawn from the supply is then found, and finally the working is unwound back through the network to give the potential difference and current for each resistor.
- 2
Why is the combined resistance of two resistors in parallel always smaller than the smaller of the two?
Model answer
Adding a second path gives the charge another route between the same two points, so a larger total current flows for the same potential difference and the effective resistance is therefore lower.
- 3
Calculate the combined resistance of a 30 Ω resistor and a 60 Ω resistor connected in parallel.
Model answer
Since 1/R = 1/30 + 1/60 = 3/60, the combined resistance is 20 Ω.
- 4
A 6.0 V supply of negligible internal resistance is connected to a 20 Ω resistor in series with a parallel combination of 30 Ω and 60 Ω. Calculate the current drawn from the supply.
Model answer
The parallel pair is equivalent to 20 Ω, so the total resistance is 40 Ω and the current is 6.0 ÷ 40 = 0.15 A.
- 5
In a circuit of a 20 Ω resistor in series with a parallel pair of 30 Ω and 60 Ω across a 6.0 V supply of negligible internal resistance, calculate the potential difference across the parallel section.
Model answer
The total resistance is 40 Ω and the current is 0.15 A, so the potential difference across the 20 Ω parallel section is 0.15 × 20 = 3.0 V.
- 6
In a circuit of a 20 Ω resistor in series with a parallel pair of 30 Ω and 60 Ω across a 6.0 V supply of negligible internal resistance, calculate the current in each of the two parallel resistors.
Model answer
The potential difference across the parallel section is 3.0 V, so the currents are 3.0 ÷ 30 = 0.10 A and 3.0 ÷ 60 = 0.050 A, and these add to the 0.15 A supplied.
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