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Sergio039 [100]
4 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?
Physics
2 answers:
Oxana [17]4 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]4 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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A car accelerates from 0 m/s to 20 m/s in 5 seconds,
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An automobile traveling along a straight road increases its speed from 72 ft/s to 84 ft/s in 180 ft. if the acceleration is cons
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<em>ANSWER: 2.3 s</em>
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When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming
Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf \varepsilon is

(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }

and since from Ohm's law

\varepsilon  = IR,

then equation (1) becomes

(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.

(a).

We are told that the change in magnetic flux is \Phi_B = 30Tm^2,  the current induced is I = 40A, and the resistance of the wire is R = 2.5\Omega; therefore, equation (2) gives

(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

which we solve for \Delta t to get:

\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},

\boxed{\Delta t = 0.3s},

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been I =20A, then equation (2) would give

(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},

\boxed{\Delta t = 0.6 s\\}

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux \dfrac{\Delta \Phi_B}{\Delta t} is greater than in part b, and therefore , the current in (a) is greater than in (b).

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