A capacitor discharging question from a TPR passage

[Maxxxx]

I understand why choice A is correct. It's because of the difference between power and energy. But can anyone explain to my why D is not also correct? I always thought in a parallel circuit, when you change the resistance of a resistor the power of the other resistor would stay the same. Since P=V^2/R. Is there something different going on when it's a discharging capacitor?

Please see the question below...

Figure 4 shows a schematic diagram of a camera’s electronic flash mechanism. It consists of a charging circuit, a discharging circuit, and an indicator circuit. When Switch S1 is closed, the battery charges the capacitor through the resistor R. Connected in parallel to the capacitor is a large resistance R1 in series with a light-emitting diode (LED), which is a circuit element that “glows” when a small voltage is applied across it. The LED turns on when the capacitor is fully charged, indicating that the flash is “ready.” When Switch S2 is depressed, the charging circuit is opened, and the discharge loop is closed. Since r, the internal resistance of the bulb, is very small, a large current surges through the flash bulb as the capacitor discharges, resulting in a brief, brilliant flash of light.

Figure 4

Camera’s electronic flash mechanism

Which one of the following is NOT an advantage of making r, the internal resistance of the bulb, as small as possible in the circuit in Figure 4?

A.
A larger r would mean less energy is delivered to the flash bulb when the capacitor discharges.

Correct Answer

B.
The discharge time constant would be larger if r were larger.

C.
The flash produced would be less bright if a larger r were used.

D.
If r were larger, more power would be lost through R1.
Explanation:
A. The energy delivered to the bulb depends on the amount stored in the capacitor. Making r smaller only changes how fast the capacitor discharges, that is, how fast this energy is delivered to the bulb (and to R1). Thus, making r smaller increases the power delivered to the bulb (since a smaller rmeans a smaller time constant). Since the same energy is transferred to the bulb in a shorter amount of time, the flash will be brighter. This is one advantage of the shorter discharge time constant. Another is given in choice D; since the bulb and the branch of the circuit containing R1 are in parallel when S2 is depressed, the smaller r is, the larger the portion of discharge current flowing through it, and thus the smaller the current through R1. This means that less of the energy gets dissipated through R1 and more is available for production of light in the bulb.

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[Maxxxx]

anyone?

 

[butters221]

This is tricky because you are right the answer A and D are both advantages. However, the advantage of D is negligible compared to A. The purpose of the circuit is to generate a flash from the bulb, not to be as energy efficient as possible. If the r was increased, the energy in the bulb would decrease, thus making the circuit less useful in its function. It is true that more power would be dissipated in R1, but this is not as significant of an advantage. Evidence for this can be inferred, because if the circuit designer was really concerned about the small amount of power lost through R1, then they would have made the switch break that loop as well so that all of the current could be diverted to the light bulb.

 

[Maxxxx]

This is tricky because you are right the answer A and D are both advantages. However, the advantage of D is negligible compared to A. The purpose of the circuit is to generate a flash from the bulb, not to be as energy efficient as possible. If the r was increased, the energy in the bulb would decrease, thus making the circuit less useful in its function. It is true that more power would be dissipated in R1, but this is not as significant of an advantage. Evidence for this can be inferred, because if the circuit designer was really concerned about the small amount of power lost through R1, then they would have made the switch break that loop as well so that all of the current could be diverted to the light bulb.

I'm not sure I follow you... sorry! It's a pretty weirdly worded question, so I'm not too worried about why my answer is wrong. I just want to make sure my concept is right.

So in the question, if the resistance of "r" is increased, what would happen to R1? I think nothing would change. Because voltage is the same in parallel circuit, so the I through R2 would not change. Am I right?

 

[butters221]

Sorry I confused you, I guess I misunderstood your question and I also misused the term power - I meant energy.
I also misunderstood their question - I think they intended to give 4 choices that do happen and we are supposed to decide which are advantages/disadvantages. It looks like they mistakenly used the term "power" instead of "energy," making an impossible question.
Did you mean R1 when you wrote R2? If so you are correct.

 

[Maxxxx]

Yes, I meant R1. Would my statement be true even if the voltage source is transient, as in the discharging capacitor case? Do V=IR and P=IV hold true even if voltage is changing?

 

[butters221]

Yes, those equations always hold true. Voltage decreases during the discharge. Since resistance is unchanged, current decreases (through each resistor) as well. Therefore power decreases over time in both loops.

 

[Maxxxx]

Thanks!