Question

Six batteries, each of 2 volts are connected in series so that they are aiding each other. Internal resistance of each cell. They are being charged using a direct supply of 110 volts. To control the current a resistor is used in series. Power drawn from the supply is​

Answer 1

Given :

The rating of each six battery = 2 volts

The rating of each internal resistance = r = 0.1 Ohm

The terminal supply voltage = V = 110 volts

To Find :

Power drawn from the supply

Solution :

As The circuit connection is series

So, Equivalent resistance = $r_e_q$  = $r_1+r_2+r_3+r_4+r_5+r_6$

                                                    = 6 × 0.1 Ohm = 0.6 Ohm

And

Equivalent battery = E =  $r_e_q$  = $E_1+E_2+E_3+E_4+E_5+E_6$

                                                 = 6 × 2 volt = 12 volt

Now, From circuit equation

 V = E + i × $r_e_q$

Or,  110 v - 12 v = i × 0.6   Ohm

Or,   98 v = i × 0.6 Ohm

∴           i = $\dfrac{98}{0.6}$

i,e  Current flow in circuit = i = 163.3  Amp

Again

∵   Power drawn = current × resistance

                           = i × $r_e_q$

                           = 163.3 Amp × 0.6 ohm

                           = 97.98    watt

Hence, The power drawn form the supply is 98 Watts    Answer