Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer 1

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300$\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery  is  $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is  $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

    The  open circuit voltage is  $V = 40.0V$

     The internal resistance is $R = 2 \Omega$

     The emf of each battery is $e = 6.00 V$

      The internal resistance of the battery is  $r = 0.300V$

      The  charging current is  $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is  $R_z$

 So this implies that

        The total resistance in the circuit is

                              $R_T = R + 2r +R_z$

Substituting values

                             $R_T = 2.6 +R_z$

And  the difference in potential in the circuit is  

                         $E = V -2e$

                 =>   $E = 40 - (2 * 6)$

                        $E = 28 V$

Now according to ohm's law

            $I = \frac{E}{R_T}$

Substituting values

           $4 = \frac{28}{R_z + 2.6}$        

Making $R_z$ the subject of the formula

So    $R_z = \frac{28 - 10.4}{4}$

           $R_z = 4.4 \Omega$

The  increase rate of   internal energy at the supply is mathematically represented as

        $Z_1 = I^2 R$

Substituting values

     $Z_1 = 4^2 * 2$

     $Z_1 = 32 W$

The  increase rate of   internal energy at the batteries  is mathematically represented as

         $Z_2 = I^2 r$

Substituting values

         $Z_2 = 4^2 * 2 * 0.3$

         $Z_2 = 9.6 \ W$

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        $Z_3 = I^2 R_z$

Substituting values

       $Z_3 = 4^2 * 4.4$

      $Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         $C = 2 * e * I$

Substituting values

       $C = 2 * 6 * 4$

      $C = 48 W$