Tag Archives: Current Electricity
How to connect the variable resistor (rheostat)
The variable resistor (rheostat) is a device to control the size of the current by adjusting the resistance on the variable resistor. Hence just device can be used to adjust the brightness of the bulb or the fan speed.
Finding unknown resistor R and setting up the electrical circuit
To find the unknown resistor R, the following apparatus are setup.
Refer to the video below for the setting up of the apparatus.
Why do you need a variable resistor (rheostat)?
- Without the variable resistor, you will have only one set of current I and potential difference V readings. Using the formula R = V/I, you are able to find the unknown resistor. But this method is not so accurate.
- Hence, to make it more accurate, we include a variable resistor to control the size of the current through the circuit. Thus having different readings of the potential difference V across the unknown resistor.
- Instead of just one set of readings of I and V, we now have about 5 sets.
- This allows us to plot a graph of V against I.
- By finding the gradient of the best fit line, we are able to find the resistance more accurately. [gradient = V / I = R, hence the gradient of V-I graph represents resistance R]
For pure metallic conductor, like the fixed resistor R, it obeys the Ohm’s Law, hence it is an ohmic conductor.
From the graph, the current I flowing the conductor is directly proportional to potential difference V across the conductor, provided physical conditions like temperature remains constant. [the graph is a straight line with constant gradient, and passes through the origin]
Rules of Series and Parallel Circuits
Rules of Series and Parallel Circuits
To understand direct current (DC) circuits, the best way is to think in terms of river system.
Series Circuit
Parallel Circuit
Series and Parallel Circuit
Examples:
Finding effective resistance when a wire is bent into different shapes
SP N2007 P1 Q16 – Which events will cause the fuse to blow?
An electric cable contains three wires live, neutral and earth. The cable is correctly wired to a plug which contains a 3A fuse. The insulation becomes damaged and bare metal wires show.
Five possible events can occur.
- A person touches the earth wire.
- A person touches the neutral wire.
- A person touches the live wire.
- The live wire touches the neutral wire.
- The live wire touches the earth wire.
How many of these five events cause the fuse in the plug to blow?
A 1 B 2 C 3 D 4
Solutions: Option B
Consider the five events:
- A person touches the earth wire – As the person is at 0V, same as the earth wire, there will be no current flowing through the person. So current through the circuit will not be affected, which is lower than the 3A fuse rating. Fuse will not blow.
- A person touches the neutral wire.- As the person is at 0V, same as the neutral wire, there will be no current flowing through the person. So current through the circuit will not be affected, which is lower than the 3A fuse rating. Fuse will not blow.
- A person touches the live wire. – The live wire is at high potential of 240 V. The person will get an electric shock. But a common misconception is that if a person gets an electric shock, the current flowing through him is very large, which is wrong. In fact, the current is very small, much smaller than the fuse rating. Assuming the average body resistance of the person is 100 000 ohms, and the potential difference in Singapore is 240 V, since I = V/R = 240/100 000 = 0.0024 A, which is lower than 3A fuse rating. Hence the fuse will not blow.
- The live wire touches the neutral wire. – This will create a short circuit as a large current which exceeds the fuse rating will from the live (240 V) to the neutral wire (0V) as that path has very low resistance. The fuse will blow.
- The live wire touches the earth wire. – This will create a short circuit as a large current which exceeds the fuse rating will from the live (240 V) to the earth wire (0V) as that path has very low resistance. The fuse will blow.
Current Electricity
Solutions: Option B
Whatever that goes in has to come out. With this basic concept, you will be able to solve this question easily.Ignoring the unknown branch,Total current going into O = 8A + 6A = 14ATotal current going out of O = 12A + 4A = 16AHence in order to be equal, 2A must be flowing in direction PO.Hence answer is B.
N2007P1Q27 – Current Electricity – Brightness of bulbs
X and Y are lamps with filaments made of the same material. The filament of lamp X is thicker and shorter than that of lamp Y. Each lamp is connected to the mains and switched on. Which is the brighter lamp and which lamp has the larger resistance?
Brighter lamp | Larger resistance | |
A | X | X |
B | X | Y |
C | Y | X |
D | Y | Y |
Solutions: Option B
Using the formula, R = ρl/A (ρ: resistivity (Ωm), l: length (m) and A: cross-sectional area (m2) Since both are of the same material, resistivity ρ is the same. X is thicker (cross-sectional area) and shorter, X will have lower resistance than Y.
Hence Y has larger resistance. Both are then connected to mains.
You can assume they are connected individually as shown below. Potential difference across each bulb is the same as the mains.
Since brightness depends on power, using P = IV, bulb X is brighter.