Physics, 06.05.2020 21:39 woodsjnjoseph3

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?

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Answer from: queenkimm26

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?

The additional resistance is $R_z = 4.4 \Omega$

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

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

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

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

Answer from: Quest

show the diagram then i will answer

Answer from: Quest

v=ir = 3x15 =45 v

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