(III) A 120-V rms 60-Hz voltage is to be rectified with a full...

Question

(III) A 120-V rms 60-Hz voltage is to be rectified with a full-wave rectifier as in Fig. 40–40, where R = 24 kΩ, and C = 35 μF. (a) Make a rough estimate of the average current. (b) What happens if C = 0.10 μF? [Hint: See Section 26–5.]

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with circuits. This problem says a student constructs a full wave bridge rectifier circuit powered by a 110 volt R MS 60 Hertz ac source. The circuit includes a load resistance of 10 kg and uses a smoothing capacitor of 47 micro Ferras. There are two parts of this problem for part one approximate the average output current. And for part two, what would be the approximate average current if 0.20 micro ferre capacitor was used. Instead, we're given a hint that says to ignore the voltage drops across the diet. Below the question we're given a circuit diagram of what was described in the problem we're asking on four possible choices as our answers or choice. A for part one, we have 2.2 milliamps. For part two, we have 7.8 milliamps or choice B. For part one, we have 2.2 milliamps. And for part two, we have 9.2 milliamps. Choice C for part one, we have 16 milliamps for part two, we have 7.8 milliamps. And for choice D for part one, we have 16 milliamps. And for part two, we have 9.2 milliamps. Now, for part one, we need to determine the average output current for this um rectifier. And in order to do that, I need to figure out how effective our smooth capa capacitor actually is. So to do that, I need to compare our effective period for our fully rectified signal as well as the time constant. So that's where we, where we'll start. So first we'll calculate our time constant. And so we call for a time constant between a resistor and capacitor. That's gonna be Tau is equal to R multiplied by C. And we have values for both our resistance and our capacitance since our is our resistance and C is our capacitance still have TAU is equal to or our resistance. That is the 10 kg. So that is going to be 10, multiplied by 10 to the three ohms. We need to convert that from kilo ohms to ohms by multiplying by 10 to the three. Then that's gonna be multiplied by our capacitance, which is the 47 micro Ferras. So I need to convert micro ferris to ferrets. So I'll have 47 multiplied by 10 to the negative six fads. Recall to convert from micro ferrous to ferrets. You need to multiply by 10 to the negative six. So plugging those values to our calculator will get Tau it's equal to 0.47 seconds. Now, the next thing we need to calculate is our period. And in the problem, we were only given a frequency for our voltage source, that was the 60 Hertz. So we need to recall our relationship between our period and frequency to recall that our period T is equal to one divided by our frequency. F. Now, we do have to be careful about the value of our frequency that we plug in. So we're gonna have T is equal to one divided by. And here, since we have a fully rectified signal, that's basically taking our sinusoid away of this coming in and taking the absolute value of it. So that effectively doubles the frequency. So for our frequency, we're gonna have two multiplied by 60 Hertz. And so when I plug those values into my calculator, I'll get T is equal to 0.008 three seconds. And here I've just kept two significant figures. So if I compare those two values, I'll see that our time constant tau much, much greater than our period T. So that means that once the signal reaches its peak voltage, it's not really going to decay much since the capacitor will always be able to supply um the voltage that is dropped due to the signal going back down to zero before coming back up to its peak again. So that means that our peak voltage is just gonna be equal to our output voltage. So we can use our peak voltage from our power supply to determine the output average current. So for this, we're actually going to use Ohm's Law. So we call from ohms Law that are current I and this is gonna be our average current which will label as I A and that's going to be equal to the voltage. In this case, it's gonna be our peak voltage which will label as VP divided by our resistance. R. Now, our peak voltage is not what we were given the problem. We are actually given the R MS voltage but recall that we do have a relationship between the peak voltage and the R MS voltage. So we have VP, it's going to be equal to VR MS multiplied by the square root of two or VR MS is our R MS voltage and VP is our peak voltage. So we can substitute that back in to our equation for the current. So we'll have our average current I A equal to VR MS multiplied by the square root of two divided by R. And so we can plug in values for um our voltage as well as our resistance. So we will have I A is equal to VR MS that was given in the problem that is 110 volts that's gonna be multiplied by the square root of two and divided by our resistance, which was 10 multiplied by 10 to the three ohms. And so plugging those values to our calculator will get I A is equal to 1.6 multiplied by 10 to the negative two amps. And here I just kept two significant figures. However, my answer choices are in milliamps. So I'll need to convert this from amps to milliamps by multiplying by the conversion factor of 1000 milliamps divided by one. And so recalculating, our current will have I A is equal to 16 milliamps. That's gonna be my answer for part one. Now, for part two, we basically have the same set up with a different capacitor. So we're gonna follow similar steps. So for part two, we again need to find the average current, but we're gonna have a new time concept. So we're gonna need to recalculate that. So Tau is still gonna be equal to RC. So we'll have Tau, it's gonna be equal to, we have the same resistance, which is the 10 multiplied by 10 to the three ohms. But now I'm multiplying this by a different capacitor and our new capacitor is 0.2 micro fair adds. So multiply that by 0.2 multiplied by 10 to the negative six fair adds. So that means that our time constant ta when I plug those values to my calculator is gonna be equal to 0.002 seconds. Now our period is going to remain the same since we still have the same voltage source. So our period T is gonna be equal to 0.0083 seconds. So now if I actually compare our time constant with our period, we see that our period T is approximately equal to for TAU. Now we call for an RC circuit that your current actually decays exponentially. And so that means that our current is gonna be equal to e raised to the minus four multiplied by our maximum current. So that means that our current is basically 2% of its maximum. So effectively, the current has gone to zero within this time period. And so that means that our voltage will also decay down to zero. So for our average voltage, it's just gonna be half of our peak voltage. And so we can use that to calculate our new average current. So our average current I A prime is gonna be equal to our peak voltage from our power source which is VP divided by. But here it's gonna be half of that peak. We'll divide that by two R. We again make the substitution that VP is equal to VR MS multiplied by the square root of two. So we'll have I A prime is equal to VR MS multiplied by the square root of two divided by two R. And we again can plug in values for those quantities. So we'll have I A prime is equal to 110 volts multiplied by the square root of two divided by the quantity of two multiplied by 10 multiplied by 10 to the three ohms. So now when I plug those values into my calculator, I'll get I A prime is equal to 7.8 multiplied by 10 to the negative three amps. And again, I'll need to convert that into milliamps. So I'll multiply that by the conversion factor of 1000 milliamps divided by one amp. And so doing so I'll get I A prime is equal to 7.8 mills. And I've kept just two significant figures in this calculation. So that's gonna be my answer for part two. So now I have both parts calculated. I can look at my answer choices and the correct answer choice corresponds to answer choice C. So the trick to this problem was to basically determine the effectiveness of the smoothing capacitor. So in part one, it basically kept the voltage at the peak voltage of the power supply. And so we could use that along with Ohm's law to calculate our average current. Now, in part two, the smoothing capa capacitor was not very effective and allowed the voltage and the current to drop down to zero. So that meant that our average voltage was just our peak voltage divided by two. And using that value for our average voltage, we could calculate our average current using Ohm's law again. So I hope that this has been useful and I'll see you in the next video

(III) A 120-V rms 60-Hz voltage is to be rectified with a full-wave rectifier as in Fig. 40–40, where R = 24 kΩ, and C = 35 μF. (a) Make a rough estimate of the average current. (b) What happens if C = 0.10 μF? [Hint: See Section 26–5.]

Key Concepts

RMS Voltage

Full-Wave Rectification

Capacitance and Filtering

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