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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