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WITCHER [35]
3 years ago
6

The potential difference between the terminal of an electrical heater is 60v when it draw a current 4A fron the source. What cur

rent will the heater draw if the potential difference is increased to 120v?
Physics
1 answer:
insens350 [35]3 years ago
6 0

The current through the heater is 8 A if potential difference is increased to 120 V.

Explanation:

  • According to Ohm's Law, the potential difference (Voltage) is directly proportional to the current flowing through a conductor.

That is, V ∝ I

⇒ V = IR where R is the constant of proportionality called resistance.

  • In this case, given voltage = 60 V and current = 4 A. To find the current, first find the constant of proportionality, resistance of the heater.

R = V/I = 60/4 = 15 ohm

  • If the potential difference is increased to 120 V, find the current flowing through it.

I = V/R = 120/15 = 8 A

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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
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Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of char
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Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

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We have to find the net flux through the closed surface.

Net electric flux,\phi=-2 EA=-2E(\pi r^2)

\phi=-2\times 300\times (3.14\times (0.66)^2)

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c.

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Q=-7.26nC

Where 1nC=10^{-9}C

7 0
3 years ago
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