## Maximum Power Transfer Theorem

In a circuit, for what value of the external resistance would the maximum power be drawn from a battery?

Alt text : maximum power transferFor the shown network, power developed in resistance R equals

P = E

_{2}R / (R+r)^{2}(l = R / R + r and P = l

^{2}R)Now, dP/dR = 0 (for P to be maximum dP / dR = 0)

Þ E

^{2}. (R + r)^{2}– 2(R) (R + r) / (R + r)^{2}= 0Þ (R + r) = 2R

Or R = r

Þ The power output is maximum when the external resistance equals the internal resistance, i.e. R = r.

## Illustration:

A copper wire having a cross-sectional area of 0.5 mm

^{2}and length of 0.1 m is initially at 25^{o}C and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire,(a) find the time in which the wire starts melting. [The change of resistance of the wire with temperature may be neglected.]

(b) What will this time be, if the length of the wire is doubled?

[Density of Cu = 9 × 10

^{3}kg m^{–3}, specific heat of Cu = 9 × 10^{–2}cal kg^{–1}^{o}C^{–1}, M.P. (Cu) = 1075^{o}C and specific resistance =1.6 × 10

^{–8}Wm].

Solution:(a) Mass of Cu = Volume × density = 0.5 × 10

^{–6}× 0.1 × 9 ×10

^{3}= 45 × 10^{–5}kg.Rise in temperature = q = 1075 – 25 = 1050

^{o}C. Specificheat = 9 × 10

^{–2}kg^{–1}^{o}C × 4.2 JÞ l

^{2}Rt = mSqÞ t = mSq/l

^{2}.RBut R = pL / A = 1.6 × 10

^{–8}× 0.1 / 0.5 × 10^{–8}= 0.558 s(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, the wire will start melting in the same time.