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 transfer

For the shown network, power developed in resistance R equals

P = E₂R / (R+r)²

(l = R / R + r and P = l²R)

Now, dP/dR = 0 (for P to be maximum dP / dR = 0)

Þ E². (R + r)² – 2(R) (R + r) / (R + r)² = 0

Þ (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 mm² and length of 0.1 m is initially at 25^oC 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 × 10³ kg m^–3, specific heat of Cu = 9 × 10^–2 cal kg^{–1 o}C^–1, M.P. (Cu) = 1075 ^oC and specific resistance =

1.6 × 10^–8 Wm].

Solution:

(a) Mass of Cu = Volume × density = 0.5 × 10^–6 × 0.1 × 9 ×

10³ = 45 × 10^–5 kg.

Rise in temperature = q = 1075 – 25 = 1050 ^oC. Specific

heat = 9 × 10^–2 kg^{–1 o}C × 4.2 J

Þ l²Rt = mSq

Þ t = mSq/l².R

But 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.