A copper wire that is 1 mm thick and 30 cm long is connected to a 1 v battery. (the resistivity of copper is 1.69 x 10-8 ω⋅m). The power is 154.82 W.
As we know the formula of Power:
P = V²/R ......(i)
Where,
P = power dissipated, V = voltage R = resistance
Now,
R = Lρ/A .....(ii)
Where,
L = length of the wire, A = cross-sectional area, ρ = resistivity of the wire.
From eq (i) and eq(ii),
P = V²/(Lρ/A)
P = AV²/Lρ........ (iiI)
Here,
A = πd²/4, Where d = 1 mm
A = (3.14×0.001²/4) = 7.85×10⁻⁷ m²
Putting the value of A and the given values from the question in equation (iii).
P = (7.85×10⁻⁷×1²)/(0.3×1.69 x 10⁻⁸)
P = 154.82 W
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