Two devices of rating 44 w, 220 v and 11 w, 220 v are connected in series. the combination is connected across a 440 v mains. the fuse of which of the two devices is likely to burn when the switch is on? justify your answer
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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 8than the rated current for the second device the fuse will be blown in this case.
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 8than the rated current for the second device the fuse will be blown in this case.
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