Question

$(1)the \: interna \: resistnce \: of \: battery \: is \: 0.4ohm$
$(2)the \: emf \: of \: battery \: is \: 1v$
$(3)r \: at \: which \: power \: is \: 5w \: is \: 2.5ohm$
$(4)at \: i \: = 2a \: power \: is \: 3.2w$
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Answer 1

$\huge\bigstar\underline\pink{Answer}$

★ Current i in the circuit is given as,

i = E / (R+r) .......1)

where E is EMF of battery , R is external Resistance and r is internal resistance .

At maximum power , external resistance R equals the internal resistance.

From graph, we see that , we get maximum power 5 W when the current is 5 A.

Hence we have , i2 × R = i2 × r = 5 W

if i = 5 A , then we have 25 × r = 5 W or r = 0.2 Ω

EMF of battery is obtained from eqn.(1) by substituting internal resistance value.

When power dissipation in external resistance R is maximum, we have

i = E / ( r + R ) = E / (2r)

Hence EMF E = i × (2r) = 5 × (0.4) = 2 V

Happy to help!

$\huge\pink{\ddot{\smile}}$