When a voltameter is connected across a battery which is connected with external resistance 280 `Omega` to measure its emf, it reads 1.4 V. Now when this emf is measured by potentiometer, it is measured as 1.55 V. Now if maximum power is to be spent in the external resistance then its value should be made equal to ...... .

Options

A. 30 `Omega`
B. 35 `Omega`
C. `45 Omega`
D. `60 Omega`

Answer

A

Explanation

`30 Omega`

Reading obtained in voltmeter, connected across a battery (i.e. across its terminals a and b) is,
V = IR
`therefore 1.4 = I xx 280 `
`therefore I = 5 xx 10^(-3)` A
emf of above battery as measure by potentiomete is `epsilon = 1.55 ` V.
For a discharging battery, we have,
`V = epsilon - Ir `
`therefore 1.4 = 1.55 - (5 xx 10^(-3)) r`
`therefore (5 xx 10^(-3)) r = 0.15 `
`therefore r = (0.5 )/(0.005) = 30 Omega`
Now, in order to spend maximum power in external resistance its value should be made equal to internal resistance of the battery. Hence, new value of external resistance should be made `30 Omega`.
Verification :
When `R_(1) = 280 Omega and r = 30 Omega ( " here R " gt r)`
`I_(1) = (epsilon)/(R_(1) + r) = (1.55)/(280 + 30) = 0.005 A `
Power spent in `R_(1)` is
`P_(1) = I_(1)^(2) R_(1) = (0.005)^(2) (280) = 0.007` W .... (1)
(ii) When `R_(2) = r = 30 Omega`
`I_(2) = (epsilon)/(R_(2) + r ) = (1.55)/(30 + 30) = (1.55)/(60) = 0.02583` A
Here , spent in `R_(2)` is
`P_(2) = I_(2)^(2) R_(2) = (0.02583)^(2) (30) = 0.2 W `.... (2)
(iii) When `R_(3) lt r , ` suppose `R_(3) = 20 Omega (lt 30 Omega) `
`I_(3) = (epsilon)/(R_(3) + r) = (1.55)/(20 + 30) = (1.55)/(50) = 0.031 A `
Now power spent in `R_(3) ` is ,
`p_(3) = I_(3)^(2) R = (0.031)^(2) (20) = 0.01922` w... (3)
Equation (1), (2), (3) prove that power spent in external resistance is maximum when R = r
At this time, `1 = (epsilon)/(R + r) = (epsilon)/(2r) (because R = r) `
Notes:
(i) Here maximum power spent in the circuit is `I^(2) (R + r) = I^(2) (2r) = ((epsilon)/(2r) )^(2) (2r) = (epsilon^(2))/(2r)`
(ii) Maximum power that can be generated in the battery is,
`P_("max") = (epsilon) (I_("max") ) = (epsilon) ((epsilon)/(r)) `(when R = 0 )
`therefore P_("max") = (epsilon^(2))/(r)`