N cells each of e.m.f. E & internal resistance r are grouped into sets of K cells connected in series. The (NK) sets are connected in parallel to a load of resistance R, then:
A
Maximum power is delivered to the load if K=√NRr
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B
Maximum power is delivered to the load if K=√rNR
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C
Maximum power delivered to the load is NE24r
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D
Maximum power delivered to the load is E24Nr
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Solution
The correct options are A Maximum power is delivered to the load if K=√NRr C Maximum power delivered to the load is NE24r Given that N cells are grouped into sets of K cells connected in series, So total NK rows each of emf KE and internal resistance Kr, are connected in parallel
Net emf of the circuit in a parallel setup is given by Enet=E1r1+E2r2+E3r3+.....1r1+1r2+1r3+.... ⇒Enet=KEKr×NK1Kr×NK=KE Net internal resistance, 1Rnet=1Kr×NK=NK2r ⇒Rnet=K2rN
From maximum power transfer theorem, we know that internal resistance is equal to external resistance ⇒R=K2rN ⇒K=√NRr........(1) Now, power delivered is P=I2R ⇒P=(KER+R)2×R ⇒P=K2E24R2×R=K2E24R.......(2) put the value of K in eq. (2) ⇒P=NRr×E24R ⇒P=NE24r