2023 A/L Physics Past Paper · Question 31 (English Medium)

Answer: (4) $278.6\,\text{V}$

Given information

What we know:

The supply has an r.m.s. value $V_{\text{rms}} = 200\,\text{V}$.

Each diode has a forward bias voltage $V_D = 0.7\,\text{V}$.

We use $\sqrt{2} = 1.4$.

What the question asks

We want: the peak value $V_{\text{out}}$ of the rectified voltage.

What each letter means:

$V_{\text{rms}}$ ($\text{V}$): the r.m.s. value of the supply.

$V_m$ ($\text{V}$): the peak value of the supply.

$V_D$ ($\text{V}$): the forward bias voltage of one diode.

$V_{\text{out}}$ ($\text{V}$): the peak value of the rectified voltage.

Our plan has three steps. First find the peak of the supply. Next find the total drop in the diodes that carry the current. Last take the drop away.

Relevant theory

practical usage an alternating current is measured and also expressed by its root mean square (r.m.s.) value.

For example, what is supplied by the national electric grid of Sri Lanka to the consuming community of the country is an alternating voltage of which the r.m.s. value is about 240 V.

Here, we consider only the bridge rectifier circuit. In the bridge rectifier, instead of a single diode, a bridge consisting of four diodes are connected as shown in Figure 1.29, is used.

the diodes D1 and D2 get forward biased and allow a current to flow through RL.

On both occasions the remaining two diodes (other than the two diodes which conduct) are reverse biased and therefore, do not contribute towards the flow of current.

VK = 0.7 V for silicon diodes and VK = 0.3 V for germanium diodes.

The first two quotes say that an alternating voltage is given by its r.m.s. value, and that the grid voltage in Sri Lanka is an r.m.s. value of about $240\,\text{V}$. The peak value is $\sqrt{2}$ times the r.m.s. value, and the question gives us $\sqrt{2} = 1.4$.

The next quotes say that a full-wave bridge rectifier has four diodes. In each half cycle two of them are forward biased and carry the current, and the other two are reverse biased. The last quote says that a silicon diode has a drop of about $0.7\,\text{V}$.

For example, think of a road with two toll gates on it. A car that goes through pays at both gates, and the money paid is taken off what the driver started with.

The current in the bridge also goes through two diodes, and each one takes $0.7\,\text{V}$ off the voltage.

Required equations

$V_m = \sqrt{2}\,V_{\text{rms}}$

$V_{\text{drop}} = 2\,V_D$

$V_{\text{out}} = V_m - V_{\text{drop}}$

These three are voltages, so each one is in $\text{V}$. The answer is in $\text{V}$.

Applying it to this question

$280$

$V_m = 1.4 \times 200$

$1.4$

$V_{\text{drop}} = 2 \times 0.7$

$278.6$

$V_{\text{out}} = 280 - 1.4$

$1.4 * v - 2 * d$

Reasoning and calculation

Now match it with the five options.

(1) $141.5\,\text{V}$ is about half of what we found. It does not match.

(2) $142.2\,\text{V}$ is about half of what we found. It does not match.

(3) $277.2\,\text{V}$ takes off $2.8\,\text{V}$, which is two diodes too many. It does not match.

(4) $278.6\,\text{V}$ is what we found. This matches.

(5) $280.0\,\text{V}$ takes off nothing. It does not match.

Checks

Unit check: $V_m$, $V_{\text{drop}}$ and $V_{\text{out}}$ are in $\text{V}$. Each intermediate value is a voltage, so each is in $\text{V}$, as it should be.

Does it make sense?

The output peak must be a little less than the supply peak, because the diodes take something off. $278.6\,\text{V}$ is a little less than $280\,\text{V}$.

The drop of $1.4\,\text{V}$ is small beside $280\,\text{V}$, as it should be for silicon diodes.

Conclusion

$V_{\text{out}} = 278.6\,\text{V}$

The peak value of the rectified voltage is $278.6\,\text{V}$.

The answer is (4).

The Department of Examinations gave the mark for this question to all candidates. We do not know its reason. One possible reason is that a student who treats the diodes as ideal, as the lesson on rectification does, gets $280.0\,\text{V}$, which is option (5).

The question gives a forward bias voltage for each diode, and the lesson describes only the bridge circuit. So we give the answer that follows from the question: two diodes carry the current at a time, and the answer is $278.6\,\text{V}$, option (4).

That the peak value is $\sqrt{2}$ times the r.m.s. value is stated with its formula in the Tamil and Sinhala lessons. The English lesson shows no formula for it, so there it is our reasoning from the quotes and from the $\sqrt{2} = 1.4$ in the question.

Teacher's note

The peak value is $\sqrt{2}$ times the r.m.s. value. In a bridge rectifier two diodes carry the current at a time, so take off $2V_D$.

Use the numbers the question gives you, here the $0.7\,\text{V}$ of each diode.

Exam tip