Power consumption based on voltage and current

Step 1 - the average voltage is 99.5 volts and the current is constant at 5 amps hence, the average power consumed over that period is 99.5 × 5 = 497.5 watts.

Step 2a - the sudden drop from 99 volts to 93 volts is an average of 96 volts hence, for that short period, the average power is 96 × 20 = 1920 watts. However, if the drop from 99 volts to 93 volts is due to a quick rise in current from 5 amps to 20 amps then it's a little more complex to calculate.

Step 2b - the average voltage is 91.5 volts at a current of 20 amps hence the power is 1830 watts.

Step 3 - is ambiguously written so I can't answer that one.

Instant power is voltage times current. But if any of those vary over time, then the power varies too: you can either compute the total energy consumed or the average power over a time span. The computation is basically the same as the average power is just the total energy divided by the time duration.

If you can assume that the voltage drop is linear over time, just use the average values of voltage and current and you will get an approximation of the power. That is, for your first interval, V~99.5V, I=5A, so average_power ~ 99.5V * 5A = 497.5 W.

For the second interval, V~91.5, I=20A, so `average_power ~ 91.5V * 20A = 1830 W.

The 3rd interval is easier: V=97, I=5, so power = 97 * 5 = 485W.

If your voltage drop is not linear, or if the current varies, and you do not want to do the approximation then you would need to know the exact shape of the voltages and current (linear, exponential, ...) and do an integration over time of the voltage times current. That will give you the total energy, that you can divide by the time to get the average power.

If we assume linear drops, for example, for the first interval, going from 100 to 99 in 5 seconds, the formula for the voltage is V(t)=100 - t/5, while the current is a constant I=5A. The formula for the instant power is P(t) = (100 - t/5)*5 = 500 - t.

Now you can integrate that between 0 and 5 and get: $$\int_0^5 (500 - t)dt = \left[500t - {1\over 2} t^2\right]_{t=0}^5 = 2487.5J$$

So the average power is 2487.5 J / 5s = 497.5 W. Naturally the same as with the average value.

When you say "power consumption", it's not clear what you mean. To illustrate my difficulty with your question, imagine you work for $10 per hour, for 8 hours (your wage), and then a further 4 hours at a rate of $20 per hour. At the end of the 12 hour shift, it makes sense to ask how many dollars you have, but not "how much wage" you have.

In your electrical system, energy is the commodity being transferred, measured in Joules, and analogous to dollars. Power is the rate at which that energy is transferred, measured in Joules per second, analogous to "dollars per hour". Asking "how much power" is like asking "how much wage".

You could figure out an average wage, for the 12 hour shift, or you can calculate the total money you accumulated during the shift, corresponding to "average power" or "total energy".

It's probably easier to assume you mean "total energy", and then later we can calculate what average power that corresponds to, so that's what I'll do.

You have 2 consecutive intervals of time, 10s and 20s long respectively, and in each interval you have a potentially varying amount of voltage \$V\$ across some load, and a potentially varying amount of current \$I\$ through the load.

At any instant in time, the power (in Watts, or Joules per second) being delivered to the load is the product of those two values \$P = V \times I\$, at that instant in time. Therefore power varies as time progresses, just as wages vary with time of day. Energy (in Joules), on the other hand, is being accumulated in the load, ever increasing.

Note that you also use the phrase "consuming current", which doesn't make sense. Current is a "flow" of charge. You may say that something "passes current", or "carries current". You would never say that a road "consumes cars".

You did most of the arithmetic for me, by correctly stating the average power in stage 1, that being the average voltage multiplied by the constant current:

$$ \begin{aligned} P_1 &= \frac{100V+99V}{2}\times 5A \\ \\ &= 497.5W \end{aligned} $$

Now your "income" from that period is is the product of your wage and the amount of time you worked at that wage, or in electrical terms, the product of power and time:

$$ \begin{aligned} E_1 &= P_1 \times 10s \\ \\ &= 497.5W \times 10s \\ \\ &= 4975J \end{aligned} $$

Then you describe a drop from 99V to 93V, with a current increase to 20A. This voltage drop and current rise is instantaneous, and those values are irrelevant for a transition that takes zero time to complete. You could work at a wage of one billion dollars per microsecond, and your net income would not change by one single cent if the amount of time you worked was exactly zero microseconds. I will disregard what happens in this instant of time.

Next, in stage 2, voltage falls uniformly from 93V to 90V, and current stays a constant 20A. This means that your average power is:

$$ \begin{aligned} P_2 &= \frac{93V+90V}{2}\times 20A \\ \\ &= 1830W \end{aligned} $$

This average power is delivered for 20s to the load, so the total energy delivered in this period is the product of power and time:

$$ \begin{aligned} E_2 &= P_2 \times 20s \\ \\ E_2 &= 1830W \times 20s \\ \\ &= 36,600J \end{aligned} $$

The total energy delivered to the load is the sum of energies \$E_1\$ and \$E_2\$, and that total energy was delivered over a total duration of 30s. So, on average, during phases 1 and 2, the power was:

$$ \begin{aligned} P_{AVERAGE} &= \frac{E_1 + E_2}{10s + 20s} \\ \\ &= \frac{4975J + 36600J}{30s} \\ \\ &= \frac{41575J}{30s} \\ \\ &= 1386W \\ \\ \end{aligned} $$

In summary, total energy delivered/consumed in the first 30s was 41575J, at an average rate (power) of 1386W.

Other answers calculated the power consumed the load. I won't repeat their work but will concentrate on the wasted power. Clearly there is a voltage drop across an internal resistance (\$R_S\$) of the 100V supply and in the wiring. It appears that \$R_S\$ is not constant, probably increasing with temperature.

Step 1:

At 99V, 5A out, there is a 1V drop across \$R_S\$. By Ohms Law \$R_S=1V/5A=0.2\Omega\$. Since the voltage dropped gradually, \$R_S\$ was a little lower at turn on, say \$0.1\Omega\$. So the average power dissipated by \$R_S\$ is \$P_{1ave}= 5^2(0.1)\approx2.5W\$.

Step 2:

When the 20A load kicks in:$$R_S=\frac{100-93}{20}=0.35\Omega$$ which is a little higher than the end of step 1. The power dissipated by \$R_S\$ at this point is $$P_{2a}=20^2X0.35=140W$$

At the end of the step 2:$$R_S=\frac{100-90}{20}=0.5\Omega$$ So $$P_{2b}=20^2(0.5)=200W$$ The average power dissipated by \$R_s\$ is \$P_{2ave}=270W\$

So the energy dissipated as heat is:$$P_{RsAve30s}=\frac{P_{1ave}T1+P_{2ave}T2}{T1+T2}$$ $$P_{RsAve30s}=\frac{(2.5)(10)+(270)(20)}{30}\approx180W$$

Step 3:

When the 20A switches to 5A, the voltage recovers to 97V not 99V. This means that the internal resistance has not yet cooled down. Its value is: \$R_S=(100-97)/5=0.6\Omega\$ dissipating 15W, less as it cools.

Using @Simon Fitch value \$P_{Lave}=1386W\$ then the efficiency is: $$\eta=\frac{P_{Lave}}{P_{Lave}+P_{RsAve30s}}\approx88\%$$

If \$R_S\$ does not have time to cool down the efficiency will be worse.