A battery of five lead acid accumulators ,each of emf 4 V and internal resistance 1 ohm connected in series - Physics - Current Electricity

Question

A battery of five lead acid accumulators ,each of emf 4 V and
internal resistance 1 ohm connected in series is charged by 100 V
dc source
Calxulate the following:
1)the series resistance to be used in the circuit to have a current
of 5 A

Neha Gupta · Accepted Answer

total emf on the circuit = 5×4 +100 = 120 VNEt resistance = 5×1+R = 5+RI = 5AV = I×net resistance120/5 = 5+R5+R = 54R = 49 ohm

A battery of five lead acid accumulators ,each of emf 4 V and internal resistance 1 ohm connected in series is charged by 100 V dc source. Calxulate the following: 1)the series resistance to be used in the circuit to have a current of 5 A

A battery of five lead acid accumulators ,each of emf 4 V and internal resistance 1 ohm connected in series is charged by 100 V dc source.
Calxulate the following:
1)the series resistance to be used in the circuit to have a current of 5 A