two identical resistors of 7ohm are each connected to a battery of 6v. calculate the ratio of the powers consumed by the resulting combinations with minimum resistance and maximum resistance.
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Answered by
6
1)in series
10 + 10 = 20Ω
now find the current
I=V/R
=6/20
=.3A
now power
P1=V(I)
=6×.3
= 1.8W
2)in parallel
1/R=1/R+1/R
=1/10+1/10
R =5Ω
now current
I=V/R
=6/5
=1.2A
power
P2=V×I
=6×1.2
=7.2W
Ratio
P1/P2=1.8/7.2
=.25W
10 + 10 = 20Ω
now find the current
I=V/R
=6/20
=.3A
now power
P1=V(I)
=6×.3
= 1.8W
2)in parallel
1/R=1/R+1/R
=1/10+1/10
R =5Ω
now current
I=V/R
=6/5
=1.2A
power
P2=V×I
=6×1.2
=7.2W
Ratio
P1/P2=1.8/7.2
=.25W
Answered by
4
With minimum resistance the resistors will be connected in parallel.
For minimum resistance:
1/Rt = 1/R1 + 1/R2
Rt = effective resistance
R1 = R2 = 7 ohms
Doing the substitution we have :
1/Rt = 1/7 + 1/7
= 2/7
Rt = 7/2 = 3.5 ohms
At maximum resistance the resistors are connected in series:
Rt = R1 + R2
= 7 + 7 = 14 ohms.
The formula for power is given as :
Power = V²/R
Power at Minimum resistance is given by:
P1 = 6²/3.5 = 10.29 watts
Power at maximum resistance :
P2 = 6²/14
= 36/14 = 2.57 watts
P1 : P2 is given by :
36/3.5 × 14/36
= 4/1
= 4 : 1
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