2019 A/L Physics Past Paper · Question 47 (English Medium)

Answer: (2) $\frac{V_mI_m}{4}$

What the question asks

We want:

The average power in the load over one period.

We use the graphs of $V(t)$ and $I(t)$.

Reading the options

The options are different fractions of $V_mI_m$.

We work out the energy in one period from the graphs.

Then we compare.

Relevant theory

At every moment the power is $P = VI$.

The average power over one period $T$ is the area under the $VI$ graph divided by $T$.

$$\bar P = \dfrac{1}{T}\int_0^T VI\,dt$$

Passive resistors are those resistors in which the electrical energy flowing through them gets completely converted into heat.

Required equations

$P(t) = V(t)\,I(t)$

$\bar P = \frac{1}{T}\int_0^T P\,dt$

Reasoning and calculation

The axes:

Across in both graphs: the time $t$. $T$ is one period.

Top: $V$ from $-V_m$ to $+V_m$.

Bottom: $I$ from $-I_m$ to $+I_m$.

The key points:

From $0$ to $\dfrac{T}{2}$, $V = +V_m$.

From $\dfrac{T}{2}$ to $T$, $V = -V_m$.

The current is $0$ from $0$ to $\dfrac{T}{4}$.

From $\dfrac{T}{4}$ to $\dfrac{T}{2}$ it rises from $0$ to $I_m$.

At $\dfrac{T}{2}$ it jumps to $-I_m$ and rises to $0$ at $\dfrac{3T}{4}$.

From $\dfrac{3T}{4}$ to $T$ it is $0$.

On each ramp the average current is $\dfrac{I_m}{2}$.

Each ramp lasts $\dfrac{T}{4}$.

$\frac{V_mI_mT}{8}$

$E_1 = V_m \times \frac{I_m}{2} \times \frac{T}{4}$

$E_2 = (-V_m) \times \left(-\frac{I_m}{2}\right) \times \frac{T}{4}$

$E = \frac{V_mI_mT}{4}$

This is the energy used in the load in one period.

In both parts $VI$ is positive.

No energy flows back.

$\frac{V_mI_m}{4}$

$\bar P = \frac{E}{T} = \frac{V_mI_m}{4}$

Checks

Unit check:

$V_mI_m$ has the unit $ext{V} imes ext{A} = ext{W}$.

Energy ÷ time has the unit $ext{J}\, ext{s}^{-1} = ext{W}$.

$Vm * Im / 4$

Conclusion

$\bar P = \frac{V_mI_m}{4}$

Only the two parts where $VI$ is positive give power.

Each gives $\dfrac{V_mI_mT}{8}$.

In one period that is $\dfrac{V_mI_mT}{4}$.

The average is $\dfrac{V_mI_m}{4}$.

So the answer is option (2).