# Evan's Space

## 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.

## How to read ammeter and voltmeter

In general, the ammeter and voltmeter can be read to 2 decimal place. Hence all recordings must be in 2 d.p.

Ammeter is a device to measure the size of the current. It is connected in series and it has very low resistance.

Voltmeter is a device to measure the potential difference between two points on the circuit. It is connected in parallel and it has very high resistance (no current should flow through them).

## 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.

## 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]

## 2013SPp1q17 What is the new P when V across resistor is doubled?

Solutions: C

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

## 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:

## Ratio of resistiviities of wires X and Y – effect of diameter on cross-sectional area

The resistance of two wires X and Y are in the ratio of 2 : 1, their lengths are in the ratio 1 : 2 and their diameters are also in the ratio of 1 : 2.

What is the ratio of the resistivities of wires X and Y?

A     1:2     B     1:1     C     2:1     D     4:1

Solutions: Option B

Note that when the diameter of the wire doubles, the cross-sectional area increases by 4 times.

Comparing the resistances of both X and Y as shown below, the ratio of resistivities is 1:1.

## N2010 P2Q12 Fuel gauge using variable resistor

Solutions:

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.

Solutions:

(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.

## N2007P1Q27 brightness of bulb

Solutions: Option B

As lamp X is thicker and shorter, these 2 factors make the resistance of X lower than Y.
Recall: Length increases, R increases & cross-sectional area increases, R decreases.
[R = pL/A]

As both bulbs are connected to the same mains, (assume 240V), the potential difference across the bulbs are 240 V.

Brightness of the bulb depends on power, P = IV or V2/R, since V is constant, I across X is higher due to lower R,
hence power of X is greater than power of Y, hence X is brighter.

## Physical Quantities (factors) that affect Resistance R

As temperature of a conductor increases, the resistance R increases.

If temperature remains constant,
1) Length, L, increases, R increases
2) Resistivity, ρ, increases, R increases
3) Cross-sectional Area, A, increases, R decreases

## What do you mean by bulb rating of 240 V 60 W?

Power is the rate of work done or energy conversion.

So a bulb of power 60 W means that the bulb converts 60 J of electrical energy to light energy (and heat energy) in one second.

## Thick Cables for High Power Electrical Appliances

High power electrical appliances are those which draws high current.

The thinner the wire (smaller cross-sectional area), the higher the resistance of the cable which brings the current into the electrical appliances.

More thermal energy will be generated due to this higher resistance in the cable. The cable might be too hot or catch fire.

So it is advisable to have thicker cables for high power electrical appliance as thicker cables have lower resistance.

## Modification to ammeter to measure bigger current

Sensitivity of the ammeter means that how small a current the ammeter can measure.

Example, if the original set up is 0.2 A per division, and the modified set up can measure up to 0.1 A per division, so we say the latter one is more sensitive as it is able to measure smaller change in the currrent.

## Which 2 junctions to connect to ahieve greatest resistance?

Solutions: B

Based on individual calculations, the highest effective resistance will only be achieved if the junctions are connected at Q and S.

(NOT connected diagonally at Q and S, which some students thought initially, as it will be short circuit and the resistors are redundant)

Important concepts which you have to know total effective resistance of parallel circuit. Refer to the image below.

## Which 2 junctions to connect to ahieve greatest resistance?

Solutions: B

Based on individual calculations, the highest effective resistance will only be achieved if the junctions are connected at Q and S.

(NOT connected diagonally at Q and S, which some students thought initially, as it will be short circuit and the resistors are redundant)

Important concepts which you have to know total effective resistance of parallel circuit. Refer to the image below.

## 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.

## N2007P1Q28 – Current Electricity – R of bulb increases as I increases

Voltage-current readings were obtained for different electrical components. Which readings are for a 3 V, 0.06 A torch bulb?

 A Voltage / V 0 1 2 3 Current / mA 0 6 12 18 B Voltage / V 0 1 2 3 Current / mA 0 25 45 60 C Voltage / V 0 1 2 3 Current / mA 0 20 40 60 D Voltage / V 0 1 2 3 Current / mA 0 10 20 30

Solutions: Option B

The rating of the bulb is 3V, 0.06A.

Hence only if a potential difference of 3V is provided across the bulb, the ideal current of 0.06A will flow through and the bulb will operate at its optimal brightness.

To solve this question, you have to find the resistance R of the bulb,  R = V/I = 3 /0.06 = 50Ω

As this is a filament bulb, the resistance increases as current increases. Filament bulb is a non-ohmic conductor.

Hence apply R = V/I for each component, the one which gives a increasing R will be the bulb.

A and D is out as when voltage is 3V, the current flowing through is not 60mA. C is out as the resistance is fixed at 50Ω throughout, not a characteristic of a bulb.

Hence only Option B is correct.

## 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.

## Drawing Voltage-Position Graph for a circuit

How to draw the voltage-position graph for the circuit.

Basic concepts:

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.