Question

two devices of rating 44W,220 V and 11W,220 V are connected in series. the combination is connected across a 440W mains. the fuse of which of the two devices is likely to burn when the switch is ON.? justify your answer

Answer 1

the total current flowing through the series circuit will be

I = 440v / RT

here

RT is the total resistance offered by the two devices

and

RT = R1 + R2 = V12/P1 + V22/P2 = (2202 / 44) + (2202 / 11)

so,

RT = 1100 ohms + 4400 ohms = 5500 ohms

so,

I = 440 / 5500 = 0.08A

and

the rated currents or the maximum current allowable for both devices will be

I1 = P1/V1 = 44/220 = 0.2A

I2 = P2/V2 = 11/220 = 0.05A

now

as the current supplied 'I' is greater than the rated current for the second device the fuse will be blown in this case.

Answer 2

the total current flowing through the series circuit will be

I = 440v / RT

here

RT is the total resistance offered by the two devices

and

RT = R1 + R2 = V12/P1 + V22/P2 = (2202 / 44) + (2202 / 11)

so,

RT = 1100 ohms + 4400 ohms = 5500 ohms

so,

I = 440 / 5500 = 0.08A

and

the rated currents or the maximum current allowable for both devices will be

I1 = P1/V1 = 44/220 = 0.2A

I2 = P2/V2 = 11/220 = 0.05A

now

as the current supplied 'I' is greater than the rated current for the second device the fuse will be blown in this case.