Evan's Space

Wonders of Physics


Leave a comment

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.

Advertisements


Leave a comment

Drawing Voltage-Position Graph for a circuit

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

current Elect

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.