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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.
A copper wire having a cross-sectional area of 0.5 mm2 and length of 0.1 m is initially at 25oC 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 oC–1, M.P. (Cu) = 1075 oC and specific resistance =
1.6 × 10–8 Wm].
(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 oC × 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.