0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# 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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Maximum power is delivered to the load if K=rNR
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
Maximum power delivered to the load is NE24r
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Maximum power delivered to the load is E24Nr
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C Maximum power delivered to the load is NE24rGiven 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

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Cell and Cell Combinations
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program