Question

4 cells of identical emf E, internal resistance r are connected in series to a variable resistor. The following graph shows the variation of terminal voltage of the combination with the current output :
(a) What is the emf of each cell used?
(b) For what current from the cell, does maximum power dissipation occur in the circuit?
(c) Calculate the internal resistance of each cell.

Answer 1

Answer:

Part a)

EMF of each cell is 1.4 volts

Part b)

For maximum power current through the cell is 1 A

Part c)

Internal resistance of the cell is 0.7 ohm

Explanation:

As per the graph the equation of voltage and current is given as

$V = -\frac{5.6 - 0}{2 - 0}i + c$

$V = - 2.8 i + c$

now at i = 0 we have c = 5.6 Volts

so we have

$V = -2.8 i + 5.6$

so we have

EMF of all 4 cells is 5.6 Volts

EMF of each cell is 1.4 volts

Part b)

For maximum power we have

$P = V. i$

$P = (-2.8 i + 5.6) i$

so we have

$\frac{dP}{di} = -5.6 i + 5.6$

so for maximum power its differentiation must be zero

so we have

$-5.6 i + 5.6 = 0$

i = 1 A

Part c)

As we know that slope of the graph is internal resistance of all four cells

so we have

$4r = 2.8$

$r = 0.7 ohm$

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Topic : internal resistance and terminal voltage of cell

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