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Sergio039 [100]

3 years ago

6

A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi

ng thru this resistor?

2 answers:

Oxana [17]3 years ago

7 0

Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

$Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A$

Therefore, for a 60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

bulgar [2K]3 years ago

5 0

Answer:

0.5 A

Explanation:

Current: The can be defined as the rate of flow of charge around a circuit. The S.I unit of current is Ampere(A)

Using,

P = VI..................... Equation 1

Where P = Power rating of the bulb, V = Voltage of the wall outlet, I = Current passing through the resistor.

Make I the subject of the equation

I = P/V................. Equation 2

Given: P = 60 W, V = 120 V.

Substitute into equation 2

I = 60/120

I = 0.5 A.

Hence the current passing through the resistor = 0.5 A

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3 0

2 years ago

Read 2 more answers

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

OR

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

3 years ago

Tonya is thinking about the topic presented in the text, "Do opposites really attract?" Which of her thoughts is an example of c

tigry1 [53]

tanya is dumb j j j j j j j j j jj j j j

6 0

3 years ago

Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0

3 years ago

What is one major difference between federal and unitary governments

nikklg [1K]

Answer:

In a unit government, all the powers of government are moved around in the central government where in turn in a federal government, the powers of government are divided between the center and the units.

Explanation:

I hope this helps.

5 0

1 year ago