NEET Class neet Answered
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Asked by jhajuhi19 | 28 Aug, 2020, 06:23: AM
Expert Answer
Current i in the circuit is given as,
i = E / (R+r) .....................(1)
where E is EMF of battery , R is external Resistance and r is internal resistance .
At maximum powwer , external resistance R equals the internal resistance.
From graph, we see that , we get maximum power 5 W when the current is 5 A.
Hence we have , i2 × R = i2 × r = 5 W
if i = 5 A , then we have 25 × r = 5 W or r = 0.2 Ω
EMF of battery is obtained from eqn.(1) by substituting internal resistance value.
When power dissipation in external resistance R is maximum, we have
i = E / ( r + R ) = E / (2r)
Hence EMF E = i × (2r) = 5 × (0.4) = 2 V
When current is 2 A , resistance R of external resistor is calculated from eqn.(1)
i = E / (r+R) or 2 = 2 / (0.2 +R)
we get R = 0.8 Ω from above equation
Hence power dissipation when current is 2 A , P = i2 R = 4 × 0.8 = 3.2 W
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Answers
(1) internal resistance of battery = 0.4 Ω
- False , we have calculated internal resistance of bettery = 0.2 Ω
(2) EMF of battery = 1 V
- False , we have calculated EMF of battery = 2 V
(3) R at which power is 5 W is 2.5 Ω
-False , 5W power is maximum power dissipation in external resistance . At maximum power dissipation external resistance R equals
internal resistance. Hence external resitance is 0.2 Ω when power dissipation is 5 W.
(4) At i = 2 A , power dissipation = 3.2 W
-True , we have calculated the power dissipation at i = 2 A and it is 3.2 W
Answered by Thiyagarajan K | 13 Oct, 2020, 11:07: PM
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