Such a problem arises when the values of some of its elements are unknown in a classical RC circuit. In Figure 1, these elements are circled with a dotted line - a square, which is, in fact, a "black box". We do not know \(r\) - the active internal resistance of the circuit (losses), and, accordingly, the current \(I_0 \) flowing through it. And also, we do not know the active resistance \(R \), through which the capacitor \(C \) is charged, and, accordingly, the current \(I_C \) flowing through it.

But we know the capacity of the capacitor \(C \) and the parameters that we can measure at the input and output of the circuit: current from the power source \(I \) at the beginning and end of the charge: \(I_1, I_2 \), capacitor voltage \(U \) at the beginning and end of the charge: \(U_1, U_2 \). We need to find the average current \(I_M \) consumed from the power supply during charging \(t_2 \). These values are shown in detail in graph 2, where the current from the power supply is shown in red, and the voltage across the capacitor is shown in blue.

Fig.1. Schematic diagram of an RC circuit for calculation

Fig.2. Graph explaining the known parameters of the task

To solve this problem, it is necessary to find all the missing elements, and then, using integration, find the desired answer. First, we use the classical solution to the problem of charging a capacitor \(C \) from [1]: \[\tag{1} U(t) = U_1 \mathrm{e}^{-t/\tau} + U_p (1 - \mathrm{e}^{-t/\tau}), \quad \tau = R C \] Based on this, we find the values on the capacitor at the initial and final moments of time (\(t_1, t_2\)). Since \(t_1 = 0\), then \(t_2\) is the charging time of the capacitor. Then: \[U (t_1) = U_1 \qquad (2) \] \[\tag{2} U(t_1) = U_1 \] \[\tag{3} U(t_2) = U_2 = U_1 \mathrm{e}^{-a} + U_p (1 - \mathrm{e}^{-a}), \quad a = t_2/\tau \] And from here we get the voltage of the power source. Why not enter it as a known quantity? - we will answer a little later. In the meantime, let's find its value from the previously obtained expressions: \[\tag{4} U_p = U_1 + {\Delta U \over 1 - \mathrm{e}^{-a}}, \quad \Delta U = U_2 - U_1 \] Then formula (1) will take its final form: \[\tag{5} U(t) = U_1 + \Delta U {1 - \mathrm{e}^{-t/\tau} \over 1 - \mathrm{e}^{-a}} \] Now let's move on to the current. From Figure 1 it is obvious that: \[\tag{6} I(t) = I_0 + I_C(t), \quad I_C(t) = {U_p - U(t) \over R} \] As in the case of voltage, here we also find the current values at the initial and final times (\(t_1, t_2 \)): \[\tag{7} I(t_1) = I_1 = I_0 + I_C(t_1), \quad I(t_2)= I_2 = I_0 + I_C(t2) \] From (6) and (7), by means of simple transformations, we find (8) and (9): \[\tag{8} R = {\Delta U \over \Delta I}, \quad \Delta I = I_1 - I_2 \] \[\tag{9} I_0 = I_1 - {\Delta I \over 1 - \mathrm{e}^{-a}} \] Substituting the derived expressions into formula (6), we obtain the dependence of the current on time: \[\tag{10} I(t) = I_1 - \Delta I\, {1 - \mathrm{e}^{-t/\tau} \over 1 - \mathrm{e}^{-a}} \] But for the answer you need the average value of the current, which we will receive further.

To find the average value, you need to take the integral of the current over time, and divide it by the charging time of the capacitor \(t_2 \): \[\tag{11} I_M = \frac{1}{t_2} \int \limits_0^{t_2} I(t)\, \mathrm{d}t \] After calculating this integral, and making simple transformations, we get the final formula:

Recall that here: \[ a = t_2/\tau, \quad \tau = R C = C {\Delta U \over \Delta I} \] \[\Delta U = U_2 - U_1, \quad \Delta I = I_1 - I_2 \] Such a formula is not found in classical textbooks and may be interesting for some calculations. Example.

In this case, formula (12) is greatly simplified: \[I_M \approx {I_1 + I_2 \over 2}, \quad a \ll 1 \] This is due to the fact that the function goes to a quasi-linear section. Then the average current can be found as the average value between the initial and final currents.

We may be interested in calculating a black box that contains a converter inside, for example, a step-up voltage (see the figure on the left). In this case, formula (12) also turns out to be fully operational, since \(\tau \) turns out to be independent of these parameters.

Also, the calculation is also suitable in the case when at the exit "Black box" impulse voltage. It is for these cases, when this voltage is unknown, that we did not enter \(U_p\) into the problem statement. The approach presented here differs in that for the energy calculation of such unknown structures, it is enough for us to know two values of the current entering it (before and after charging), two voltage values on the charged capacitor (before and after), the capacitance of the capacitor, and the time of its charge.

By multiplying the average current value according to formula (12) by the supply voltage B1, we get the average power consumed by the power supply when charging the capacitor. And by multiplying this power by the charging time, we find the energy spent on this process.