This simple circuit involves a resistance wire on the meter rule and the use of jockey tapping at different length L on the resistance wire.
Note that this is not a potential divider. Rather this set up works like a variable resistor (rheostat) in the circuit. When the switch is closed (jockey NOT tapping on resistance wire), there is no current flowing as voltmeter (infinity resistance) is connected in series with the ammeter and the battery. Hence the voltmeter is showing the battery’s electromotive force (emf) of 3.0 V.
When the switch is closed and the jockey is tapped on the resistance wire on the ruler, the longer the L (length of resistance wire), the higher the resistance of the circuit, the higher the potential difference (voltmeter) across the resistance wire and the smaller the current through the ammeter. Hence the tapping of the jockey on the resistance wire is similar to adjusting the resistance on the variable resistor.
(If you are wondering why the voltmeter is not showing the emf of the battery when the jockey touches the resistance wire L, it is because in reality there is internal resistance in the battery or even in the connecting copper wire. For olevel theory we assume no internal battery resistance or resistance in copper wire or ammeter).
To learn how to set up the experiment, refer to the video below.
Both ammeter and voltmeter have two terminals, (positive and negative). The conventional current must flow into the positive terminal (+) of the meters and out of the negative terminal (-). If the connection is the opposite, the needle will deflect below the zero marking. Refer to the video below.
Sometimes, the connection terminals on the voltmeter and ammeter are different. Likewise, different types of wire with different connection heads have to be used. Refer to the video to see how are they connected in general.
Solutions: Option C
[There is another similar question in 2013 Nov Pure Physics P1Q40. The answer is Option C too. Refer to the last section of this post]
Refer to the 2 videos below. The 1st one is faster if you know that the bigger the resistance, by proportion, the bigger the potential difference of the component as it will take a larger portion of the emf wrt to the other components in series. This method is especially useful for MCQ.
If you are still unsure, you may put in values to find the I, V and P across the components. This will be more tedious and time-consuming.
Another similar question is 2013 Nov Pure Physics P1 Q40.
Solutions: Option C
Answer: Option C
These are some of the necessary electrical symbols that you must know. Those in grey background are for Pure Physics.
Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.
Lighting circuit, in fact all circuits at home, should be connected in parallel.
When one bulb is spoilt or switched off, the rest of the bulbs can still function normally at normal brightness. This is because the potential difference across each bulb in the branches remains the same.
If the lighting circuit is connected in series, when one bulb is spoilt or switched off, it will be an open circuit and no current can flow through the circuit. Hence all the bulbs cannot function.
Rules of Series and Parallel Circuits
To understand direct current (DC) circuits, the best way is to think in terms of river system.
Series and Parallel Circuit
View the video on how the gauge works.
As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.
(i) Both the fixed resistor and sensor are in series.
Total effective resistance Re = 5000 + 1000 = 6000 ohms
V = IR
12 = I x 6000
I = 0.0020 A
Hence potential across Y, V = IR (where I is constant in a series circuit)
= 0.0020 x 1000
= 2.0 V
(ii) When the temperature increases and the resistance of sensor Y decreases, the total effective resistance of the circuit decreases. Current flowing through the circuit increases.
Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, the potential difference across the 5000 ohms resistor will increase.
As resistance of the rheostat increases, the total resistance in parallel will increase. Hence potential difference across the parallel bulb Q increases, therefore BRIGHTER.
Total effective resistance of the circuit (include resistance of bulb P) increases, main current thought circuit (bulb P) decreases. Hence bulb P DIMMER.
You can think in terms of p.d of bulb P. Since p.d. across parallel branches increases, p.d across bulb P decreases, hence bulb P is DIMMER.
Without calculation, using resistance and p.d. is easier. To convince yourself, refer to the following diagrams with imaginary numbers.
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An a.c. circuit is shown below.
Which of the following diagrams show the voltage V across the resistor?
Solutions: Option C
The power supply is a alternating supply, hence direction of current changes many times in 1 second. (alternating current = a.c.)
You may ask which direction of current should you first assume in order to represent the first ‘hump’?
That is not that important. Rather you must be able to know that the ‘humps’ should alternate as the V across alternates due to the diodes.
When current flows in anticlockwise, current will flow through the 6Ω and the 12Ω (in series). When the current flows in clockwise, current will flow through the 12Ω and the 12Ω (in series). Hence V across the 6Ω and the 12Ω will alternate. Hence B and D are out.
D is out as there is an continuous alternating current flowing through the circuit. Not rectification due to the diodes in the circuit.
The diodes in the 2 branches only stop current from flowing in either branch at any one time, current still continue to flow through the circuit.
Consider when current flows anticlockwise, the V is across the 6Ω. As 6Ω is smaller than the 12Ω in series, the p.d. across 6Ω is smaller. Hence a smaller ‘hump’.
On the other hand, when current flows clockwise, the V is across the 12Ω. As this 12Ω is equal to the 12Ω in series, the p.d. across will be higher than across 6Ω previously. Hence bigger ‘hump’. So C is the answer.
How to draw the voltage-position graph for the circuit.
– Electromotive force (emf) is the work done in bringing a unit charge across the whole circuit.
– Potential difference (p.d.) is the work done in bringing a unit charge across the conductor (e.g resistor, bulb)
– Sum of potential differences across the conductors in a series circuit is equal to the emf of the battery.
You can assume there is no potential drop between 2 points on a wire.