Question

A light bulb is connected to a 120.0-V wall socket. The current in the bulb depends on the time t according to the relation I = (0.644 A) sin [(394 rad/s)t]. (a) What is the frequency of the alternating current? (b) Determine the resistance of the bulb's filament. (c) What is the average power delivered to the light bulb?

Answer 1

a) Frequency: 62.7 Hz

b) Resistance: $186.3\Omega$

c) Average power: 38.6 W

Explanation:

a)

An alternating current can be written in the form

$I(t)=I_0 sin(\omega t)$

where

$I_0$ is the maximum current in the bulb

$\omega=2\pi f$ (1) is the angular frequency, where f is the frequency

t is the time

Here the current in the bulb is

$I(t)=(0.644 A)(sin (394 rad/s\cdot t)$

This means that

$\omega=394 rad/s$

And therefore by using formula (1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{394}{2\pi}=62.7 Hz$

b)

The resistance of the bulb's filament can be found by using Ohm's law:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

Here we have:

$V_0 = 120.0 V$ (maximum voltage)

$I_0 = 0.644 A$ (maximum current)

Therefore, the resistance of the bulb is

$R=\frac{V_0}{I_0}=\frac{120}{0.644}=186.3 \Omega$

c)

In this circuit, the average power delivered to the light bulb is equal to the average power dissipated by the resistor.

For an AC circuit, the average power dissipated on a resistor is given by

$P=I_{rms}^2 R$

where

$I_{rms}$ is the rms current

The rms current can be calculated as

$I_{rms}=\frac{I_0}{\sqrt{2}}$

where in this case

$I_0 =0.644 A$ is the peak current

Here the resistance is

$R=186.3 \Omega$

Therefore, the average power is:

$P=(\frac{I_0}{\sqrt{2}})^2R=(\frac{0.644}{\sqrt{2}})^2(186.3)=38.6 W$

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