Two identical immersion rod heaters are to be used to heat water in a large container which one of the following arrangement would heat the water faster ?
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Dear student,
Question is incomplete.It should have been heaters connected in series and parallel.
Effective resistance when heaters are connected in series:
R1=R2=R( identical rods)
Effective resistance Rs=R+R=2R
Current ,I= V/Rs
Is=V/2R
Amount of Heat produced=I²(s)Rst
=(V²/4R)x2R*t
Hs=( V²/2R)*t-----(1)
Effective resistance if heaters connected in parallel:
1/Rp=1/R+1/R
1/Rp=2/R
Rp=R/2
current, I= V/R
Ip=V/R
=2V/R
Heat produced =Hp=I²(p)Rpt
=4(V/R)²(R/2)xt
=(4V²/R²)(R/2)*t
=2(V²/R)*t
We can see that Hp> Hs
Hence if heaters are connected in parallel, heat produced is faster and water gets heated up faster.
Question is incomplete.It should have been heaters connected in series and parallel.
Effective resistance when heaters are connected in series:
R1=R2=R( identical rods)
Effective resistance Rs=R+R=2R
Current ,I= V/Rs
Is=V/2R
Amount of Heat produced=I²(s)Rst
=(V²/4R)x2R*t
Hs=( V²/2R)*t-----(1)
Effective resistance if heaters connected in parallel:
1/Rp=1/R+1/R
1/Rp=2/R
Rp=R/2
current, I= V/R
Ip=V/R
=2V/R
Heat produced =Hp=I²(p)Rpt
=4(V/R)²(R/2)xt
=(4V²/R²)(R/2)*t
=2(V²/R)*t
We can see that Hp> Hs
Hence if heaters are connected in parallel, heat produced is faster and water gets heated up faster.
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