A power of 36 W is to be dissipated in a register connected across the terminal of a battery, having emf of 20 V and an internal resistance of 1 ohm. Find i) What value of resistance will satisfy this condition ii) the terminal voltage of the battery for each resistance iii) the total power expenditure in each case.
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Given
P = 36 W
E = 20 V
r = 1 ohm
To find
i)What value of resistance will satisfy this condition
ii) the terminal voltage of the battery for each resistance
iii) the total power expenditure in each case.
Solution
i) We know that Power P = V x I
= (I
x R) x I
= I2
x R
Let us
consider that resistance is RL
So power
dissipated , P = IL2 x RL
36 = IL2
x RL
IL2 = 36 / RL
IL = √ (36 / RL)
-------------------------------(1)
Kirchoff’s
voltage law states that algebraic sum of potential difference in a loop = zero
Applying KVL
in the circuit
E – VL
= 0
E – IL (r
+ RL) = 0
20 – IL
(1 + RL) = 0
IL (1
+ RL) = 20
IL = 20 / (1 + RL) ---------------------------------------(2)
Comparing
and equating (1) and (2)
√ (36 / RL) = 20 / (1 + RL)
Squaring on
both sides
36 / RL
= 400 / (1 + RL)2
400 x RL
= 36 x (1 + RL)2
400 x RL
= 36 x (1 + 2RL + RL2) ------ (a+b)2 = a2+2ab+b2
400 x RL
= 36 + 72RL + 36RL2
36RL2
- 328RL + 36 = 0
By applying quadratic formula eg, ax2 + bx + c = 0, then x = (-b±√(b^2-4ac))/2a
Applying the same formula RL = (328±√(328^2-4x36x36))/(2x36)
RL = 0.111 Ω or 9 Ω
ii) For RL = 0.111 (Case 1) For RL = 9 (Case 2)
IL = √36 / RL IL = √36 / RL
IL = √36 / 0.111
IL = √36 / 9
IL = 18 A IL = 2 A
As per KVL
E – IL (r
+ RL) = 0
E – IL
r– IL RL = 0
E – IL r – V = 0
Case 1 Case
2
20 – (18 x 1) = V 20 – (2 x 1) = V
V = 2
volts V = 18 volts
iii) P = IL2 x (r + RL)
Case 1
Case 2
P = 182 x (1 + 0.111)
P = 22 x (1 + 9)
P = 359.6 W
P = 40 W
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