A batteries of emf 1 V and internal resistance 1 ohm each in

A batteries of emf 1 V and internal resistance 1 ohm each in are connected to the external resistance A 11 The power liberated by the external resistance is: 1 mW 4 W 4 kW 1 W None of the above

Solution

Emf of the single battery E = 1 Volt

Four batteries are connected in series.

So, resultant emf E \' = E + E + E + E

                              = 4 E

                             = 4 ( 1 volt)

                            = 4 volt

Internal resistance r = 1 ohm

Four internal resistances and external resistance are in series.

So, resultant resistance R \' = r + r + r + r + R

Where R = external resistance = 4 ohm

So, R \' = 1 + 1 + 1 + 1+ 4

          = 8 ohm

Current in circuit i = E \' / R \'

                           = 4 / 8

                          = 0.5 A

Power liberated by the external resistance P = i ² R

                                                                 = 0.5 ² (4)

                                                                 = 1 W