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Question
Fri August 19, 2011

# 150 cells are supplying maximum current to external resistance of 0.6 ohm.the e.m.f. of each cell is 2 volts and internal resistance is 0.1 ohm show the arrangement of cell connection and find the value of current applied

Tue August 23, 2011
for maximum cyrrent, the emf should be maximum and hence the cells are connected in series.
Therefore, emf of the battery = 150 x 2 = 300 volts
total internal resistance of the batter = 150 x 0.1 = 15 ohm.
Total resistance of the circuit = 15 + 0.6 = 15.6 ohm
Now, current = tot. emf / tot. resistance = 300 / 15.6 = 19.2 Ampere
Related Questions
Wed January 18, 2017

# A cell of emf E and internal resistance r is connected to two external resistance R1 and R2 and a perfect ammeter.The current in the circuit is measured in four different situation: (i)without any external resistance in the circuit (ii)with resistance R1 only (iii)with Ri and R2 in series combination (iv)with R1 and R2 in parallel combination The currents measured in the four cases are 0.42A,1.05A,1.4A and 4.2A,but not necessarily in that order.Identify the currents corresponding to the four cases mentioned above.

Thu November 24, 2016

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