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Allisa [31]
3 years ago
13

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

lly uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

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3 years ago
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At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?​
Ipatiy [6.2K]

The resistance of the sample is 682.5\Omega

Explanation:

The relationship between resistance of a material and temperature is given by the equation

R(T)=R_0(1+\alpha (T-T_0))

where

R_0 is the resistance at the temperature T_0

\alpha is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

R_0 = 525 \Omega when the temperature is T_0 = 20^{\circ}C

While the temperature coefficient of resistance of nickel is

\alpha = 0.006/^{\circ}C

Therefore, the resistance of the sample when its temperature is

T=70^{\circ}C

is

R=(525)(1+0.006(70-20))=682.5 \Omega

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3 years ago
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3 years ago
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0
3 years ago
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