2 identical immersion heaters are to be used to heat water, in a large container. Which one of the following arrangement would heat the water faster: connecting the heater in series with the main supply ; connecting the heater in parallel with the main supply. Give reasons fr the answer.
2 identical immersion heaters are to be used to heat water, in a large container. Which one of the following arrangement would heat the water faster: connecting the heater in series with the main supply ; connecting the heater in parallel with the main supply. Give reasons fr the answer.
Let V be the voltage applied and R be the resistance of the individual coil.
When in series, the equivalent resistance is, R s = R + R = 2R
Current drawn is, I s = V/R s = V/2R
So, heat produced in time ‘t’ is,
H s = I s 2 R s t
=> H s = (V/2R) 2 (2R) t
=> H s = V 2 t/(2R)
When connected in parallel,
The equivalent resistance is, R p = R/2
Current drawn is, I p = V/(R/2) = 2V/R
Heat produced in time ‘t’ is, H p = (2V/R) 2 (R/2) t
=> H p = 2V 2 t/R
Thus, heat produced in the parallel combination is more.
Let V be the voltage applied and R be the resistance of the individual coil.
When in series, the equivalent resistance is, R s = R + R = 2R
Current drawn is, I s = V/R s = V/2R
So, heat produced in time ‘t’ is,
H s = I s 2 R s t
=> H s = (V/2R) 2 (2R) t
=> H s = V 2 t/(2R)
When connected in parallel,
The equivalent resistance is, R p = R/2
Current drawn is, I p = V/(R/2) = 2V/R
Heat produced in time ‘t’ is, H p = (2V/R) 2 (R/2) t
=> H p = 2V 2 t/R
Thus, heat produced in the parallel combination is more.
