You have two capacitors C,-50mF and C,-19mF. First capacitors are connected to a

Question

You have two capacitors C,-50mF and C,-19mF. First capacitors are connected to a V-5V DC battery and an idealized light bulb of constant resistance R= 7 by closing switch S1 as shown in figure below with switch S2 open. When will the light bulb shine the brightest? O Time-RC1 seconds after the switch S1 was closed O Immediately right after switch s, is closed O The light bulb will not light up because capacitor is a gap in the circuit so the circuit is never closed O Long time after the switch Si was closed when the current through the bulb is constant and brightness no longer changes O The brightness of the light bulb stays constant once the switch S, is closed The second time the experiment is repeated and capacitors are connected to the battery with switch Si while switch S2 is closed. How does the peak brightness of the bulb compare in two cases? The peak brightness of O The peak brightness of the bulb will be the same in both cases because the peak power only depends on the voltage O The peak brightness of the bulb will be the same in both cases because once the light bulb is on it stays on in both O The peak brightness of the light bulb will be higher in second case because capacitors in parallel store more energy therefore more energy will be released in form of light of the battery and resistance of the light bulb cases and the current through the light bulb is determined by the voltage of the battery and bulb's resistance O In both cases the light bulb is never connected to the battery directly so the bulb wouldn't light up O The peak brightness of the light bulb will be higher in first case because the effective capacitance and thus the time constant of the first circuit is smaller so all energy will be released in a shorter amount of time

Explanation / Answer

This is a problem of analysis, the best way to do this is by having the terms that govern each each

Part A

I = E/R e-t/RC

Let's see what happens in some cases

t = 0 in this case the current is maximum

t = inf in this case the current is zero

If we analyze the different points we see that the second point is correct

immediately that the switch S1 is closed

Part B

The two capacitors being parallel

Ceq = C1 + C2

I = E/R e-t/RCeq

As discussed in the first case the higher brightness occurs at t = 0 s for which only depends on the voltage and resistance, therefore the correct answer is two

Part C

When the battery is removed capacitors are supplying the load to keep on the bulb, as each capacitor stores a charge Q1 and Q2 and the load capacitance adds more time passes so the bulb shines longer.

Consequently, the correct answer is the one

Part D

Circuito 1

I = Io e-t/RC

I/Io= 0.3

0.3 = e-t/RC

ln 0.3 = -t/RC

t = -RC ln 0.3

t = - 7 50 10-3ln 0.3

t= 0.421 s

Q= C E ( 1 – e-t/RC)

Q = 50 10-35 ( 1 – exp( -0.421/ (7 50 10-3)

Q= 0.250 ( 1 – 0,300)

Q= 0.175 C

Q = 175 mC

Circuito 2

I = Io e-t/RCeq

I/Io= 0.3

0.3 = e-t/RCeq

ln 0.3 = -t/RC

t = -RC ln 0.3

t = - 7 (50 10-3+ 19 10-3) ln 0.3

t= 0.581 s

Q= Ceq E ( 1 – e-t/RC)

Ceq = C1 +C2

Q = 69 10-3 5 ( 1 – exp( -0.581/ (7 69 10-3)

Q= 0.340 ( 1 – 0,300)

Q= 0.238 C

Q = 238 mC

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