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Wonders of Physics

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Simple electric circuit set up

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.

Click here for another set up of a simple electrical circuit to determine the unknown resistor.



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2016 Pure Physics P1 Q40 – Voltage and Power of Bulb Y when X is switched off

2016 PPp1q40

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.

2013 PP p1q40

Solutions: Option C