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.
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.
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.
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.
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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.
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.
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