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2 years ago

When a wire is heated, the current goes down from 0.3A to 0.26A. State how the resistance of the wire has changed

Physics

1 answer:

hram777 [196]2 years ago

5 0

<h2>Hello there! :)</h2>

It's very wonderful to help you today!

Answer:

The resistance of the wire after being heated is approximately 1.15 times the resistance before being heated.

Explanation:

The resistance of the wire goes up as the temperature goes up.

By Ohm's law, we have the relationship between the resistance of a conductor R, the current flowing through it, I, and the voltage across it, V,

$V=R~x~I$

As described, the current through the wire decreases so, to maintain the same voltage, the resistance must increase when the wire is heated. Let's say R1 is the resistance before heating the wire and R2 is the resistance after,

$V=R_{1} x~I_{1} =R_{2}~} x~I_{2}$

Let's solve for R2,

$R_{2} =\frac{I_{1} }{I_{2} }~x~R_{1}$

In this case, I1 = 0.3A and I2 = 0.26A,

$R_{2} =\frac{0.3A}{0.26A} ~x~R_{1} =1.15R_{1}$

Hence, the <u><em>resistance of the wire after being heated is approximately 1.15 times the resistance before being heated.</em></u>

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$