Question

A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you can assume is the time constant of the capacitor), during which it produces an average 55 mW from an average voltage of 3.1 V. 25% Part (a) How much energy, in joules, does it dissipate? E 25% Part (b) How much charge, coulombs, moves through the light? q= 1 25% Part (c) Find the capacitance of the light, in farads. C = 1 25% Part (d) What is the resistance, in ohms, of the light? R= |

Answer 1

The correct solution is:

(a)
`1.375* 10^(-2) \ J`

(b)
`4.43* 10^(-3) \ C`

(c)
`1.42* 10^(-3) \ F`

(d)
`178.57 \ \Omega`

Step-by-step explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

`P_(avg)=55 \ mW`

`=55* 10^(-3) \ W`

Average voltage,

`V_(avg)=3.1 \ V`

Now,

(a)

⇒
`E=P_(avg)* \zeta`

On substituting the values, we get

⇒
`=55* 10^(-3)* 0.25`

⇒
`=1.375* 10^(-2) \ J`

(b)

⇒
`E=Q* V_(avg)`

then,

⇒
`Q=(E)/(V_(avg))`

On substituting the values, we get

⇒
`=(1.375* 10^(-2))/(3.1)`

⇒
`=4.43* 10^(-3) \ C`

(c)

⇒
`C=(Q)/(V)`

⇒
`=(4.43* 10^(-3))/(3.1)`

⇒
`=1.42* 10^(-3) \ F`

(d)

As we know,

⇒
`R=(1)/(4C)`

⇒
`=(1)/(4* 1.42* 10^(-3))`

⇒
`=(1000)/(5.6)`

⇒
`=178.57 \ \Omega`