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

Answer:

The correct solution is:

(a) $1.375\times 10^{-2} \ J$

(b) $4.43\times 10^{-3} \ C$

(c) $1.42\times 10^{-3} \ F$

(d) $178.57 \ \Omega$

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

$P_{avg}=55 \ mW$

       $=55\times 10^{-3} \ W$

Average voltage,

$V_{avg}=3.1 \ V$

Now,

(a)

⇒ $E=P_{avg}\times \zeta$

On substituting the values, we get

⇒     $=55\times 10^{-3}\times 0.25$

⇒     $=1.375\times 10^{-2} \ J$

(b)

⇒ $E=Q\times V_{avg}$

then,

⇒ $Q=\frac{E}{V_{avg}}$

On substituting the values, we get

⇒     $=\frac{1.375\times 10^{-2}}{3.1}$

⇒     $=4.43\times 10^{-3} \ C$

(c)

⇒ $C=\frac{Q}{V}$

⇒     $=\frac{4.43\times 10^{-3}}{3.1}$

⇒     $=1.42\times 10^{-3} \ F$

(d)

As we know,

⇒ $R=\frac{1}{4C}$

⇒     $=\frac{1}{4\times 1.42\times 10^{-3}}$

⇒     $=\frac{1000}{5.6}$

⇒     $=178.57 \ \Omega$