Question

The capacitor in the flash of a disposable camera has a value of 165 μF. 1) What is the resistance of the filament in the bulb if it takes 10 s to charge the capacitor to 80% of its maximum charge? (Express your answer to two significant figures.)

Answer 1

Answer:

3.8 x 10⁴ Ω

Explanation:

C = Capacitance = 165 μF

R = resistance of the filament = ?

t = time taken to charge the capacitor = 10 s

Q₀ = maximum charge stored by capacitor

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

Charge stored by capacitor at any time is given as

$Q = Q_{o}(1 - e^{\frac{-t}{T}})$

$0.80 Q_{o} = Q_{o}(1 - e^{\frac{-10}{T}})$

T = 6.21 s

Time constant is given as

T = RC

6.21 = R (165 x 10⁻⁶)

R = 3.8 x 10⁴ Ω

Answer 2

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

What will be the resistance in the filament of the bulb?

It is given that:-

Capacitance  C= 165 μF

Resistance of the filament = R=?

Time taken to charge the capacitor = t = 10 s

Now,

Q₀ = maximum charge which can be stored by the capacitor

Since the capacitor to 80% of its maximum charge

Then,

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

The charge stored by the capacitor at any time is given as

$Q=Q_{0} (1-e^{\dfrac{-t}{T} } )$

$0.80Q_{0} =Q_{0} (1-e^{\dfrac{-10}{T} } )$

$T=6.21 sec$  

Now the Time constant is given as

$T=R\times C$

$6.21= R\times (165\times 10^{-6} )$

$R=3.8\times 10^{4}$Ω

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

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