Question

PLEASE DO ANSWER IMMEDIATELY GUYS

A series battery of 6 lead accumulators each of emf 2 volt and internal resistance of 0.5 ohm is charged by 100 volt dc supply.


what series resistance should be used in the charging circuit in order to limit the current to 8 A??Using the required resistance obtain


1) Power supplied by dc source 2) the power supplied by the d.c energy stored in the battery in 15 min

Answer 1

Answer:

1=920w

2=872w

Explanation:

1

$p = {v}^{2} \div r$

$r = (53 + 6 \times 0.25)ohm$

$(53 + 1.5)ohm$

$54.5ohm$

$\frac{230 \times 230}{54.5}$

$920w$

2

$v4 = 230 - 6 \times 2$

$230 - 12$

$218v$

$p = \frac{v \times v}{r}$

$\frac{215 \times 215}{54.5}$

$872w$

Answer 2

Explanation:

Given, Emf of the storage battery E = 8V.

Given Internal resistance r = 0.5.

Given DC supply voltage V = 120.

Given Resistance R = 15.5.

We know that:

I = Effective voltage/R + r

  = (V - E)/(R + r)

  = (120 - 8)/(15.5 + 0.5)

  = 112/16

  = 7A.

The terminal voltage of the battery during charging:

V = E + Ir

    = 8 + 7 * 0.5

    = 11.5V.

Purpose of series resistor:

= > Resistor in series has a common current flowing through them.

= > It controls the current drawn from external supply!