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xxTIMURxx [149]

4 years ago

13

A constant voltage is applied across a circuit element. If the resistance of the element is doubled, what is the effect on the p

ower dissipated by this element? The power dissipated is reduced by a factor of 2. The power dissipated remains constant. The power dissipated is doubled. The power dissipated is quadrupled. The power dissipated is reduced by a factor of 4.

1 answer:

miv72 [106K]4 years ago

5 0

Answer:

The power dissipated is reduced by a factor of 2

Explanation:

The power dissipated by a resistor is given by:

$P=I^2 R$

where

I is the current

R is the resistance

by using Ohm's law, $I=\frac{V}{R}$ , we can rewrite the previous equation in terms of the voltage applied across the resistor (V):

$P=(\frac{V}{R})^2R=\frac{V^2}{R}$

In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have $R'=2R$ while V remains the same; substituting into the formula, we have:

$P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}$

so, the power dissipated is reduced by a factor 2.

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$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

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$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

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$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

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$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

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