Question

The current through a lamp connected across 120 v is 0.75 a when the lamp is on. (a) what is the lamp's resistance when it is on? (b) when the lamp is cold, its resistance is one fifth as great as it is when the lamp is hot. what is the lamp's cold resistance? (c) what is the current through the lamp as it is turned on if it is connected to a potential difference of 120 v? a

Answer 1

(a) The relationship between the potential difference in a resistor, the current through it and the resistance is given by Ohm's law:

`V=IR`
where
V is the potential difference
I is the current
R the resistance

Re-arranging the equation, we can calculate the resistance of the lamp when it is on:

`R= (V)/(I)= (120 V)/(0.75 A)=160 \Omega`


2) When the lamp is hot, its resistance is
`160 \Omega`. When it is cold, its resistance is one fifth of this value, therefore:

`R_c = (1)/(5)R= (1)/(5)(160 \Omega)=32 \Omega`


3) I assume here the problem is asking for the current through the lamp when the resistance is the value found in part 2) (otherwise the current is the same as point 1). To solve this part of the problem, we can apply again Ohm's law, so we find the new current:

`I= (V)/(R_c)= (120 V)/(32 \Omega)=3.75 A`