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gogolik [260]
4 years ago
12

The current through inductance L is given by I=I0e−t/τ

Physics
1 answer:
ohaa [14]4 years ago
6 0
ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T

b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V

E. t = ms
ΔVl = 0.032V
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8 0
4 years ago
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Find the ratio of the diameter of aluminium to copper wire, if they have the same
kicyunya [14]

Answer:

1.24

Explanation:

The resistivity of copper\rho_1=2.65\times 10^{-8}\ \Omega-m

The resistivity of Aluminum,\rho_2=1.72\times 10^{-8}\ \Omega-m

The wires have same resistance per unit length.

The resistance of a wire is given by :

R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}

According to given condition,

\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

3 0
3 years ago
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Answer:

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3 years ago
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solniwko [45]

Answer:

D: 0.239

Explanation:

Equation for ideal efficiency is;

η = 1 - (T_c/T_h)

We are told that;

steam comes out at 112° C. Thus, T_h = 112°C. Converting to Kelvin gives; T_h = 112 + 273 = 385 K

The one exiting into the condenser is kept at 20°C. Thus; T_c = 20 + 273 = 293 K

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