# RMS current and power calculations for a triangle waveform

## Derive an expression for the signal over one period

We starting by noticing that the period of this signal is : $$T = 20$$ We now want to come up with an expression that defines the waveform over the course of one period. If we look at the waveform starting at t=5 going till t=15 we see that is the graph of the line i=-6t (slope = -6) shifted to the right 10 units. We express this using the simple equation of a line: $$i(t) = -6(t-10)$$ $$i(t) = -6t + 60 \qquad, 5 \lt t \lt 15$$ If we look at the waveform from t=15 till t=25, we see that it is the graph of the line i=6t (slope=6) shifted 20 units to the right. Therefore we have: $$i(t) = 6(t-20))$$ $$i(t) = 6t-120 \qquad,15 \lt t \lt 25$$ We now have the following piece-wise expression for i(t) over one period: $$i(t) = \left\{ \begin{array}{ll} -6t+60 & 5 \lt t \lt 15 \\ 6t-120 & 15 \lt t \lt 25 \end{array} \right.$$

## Calculate RMS current:

Recall the generic expression (derived in the previous page)for the RMS value of a periodic function x(t): $$X_{rms} = \sqrt{\frac{1}{T} \int_0^T x^2 \; dt}$$ ...and apply it to our signal: $$I_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2 \; dt}$$ Since our waveform is expressed as a piece-wise function containing two parts, we will be evaluating two integrals (one for each half of the period): $$I_{rms} = \sqrt{\frac{1}{20} \Big[ \int_5^{15} (-6t+60)^2 \; dt + \int_{15}^{25} (6t-120)^2 \Big] \; dt}$$

$$= \sqrt{\frac{1}{20} \Big[ \int_5^{15} 36t^2-720t+3600 \; dt + \int_{15}^{25} 36t^2-1440t+14400 \Big] \; dt}$$ $$= \sqrt{\frac{1}{20} \Big[ (12t^3-360t^2+3600t)\Big|_5^{15} + (12t^3-720t^2+14400t)\Big|_{15}^{25} \Big] }$$

$$=\sqrt{\frac{1}{20} \Big[ (13500-10500) + (97500-94500) \Big]}$$ $$=\sqrt{\frac{1}{20}(6000)}$$

$$I_{rms} = 17.32 \;A$$

## Calculate power delivered to a 12 ohm resistor:

Recall that towards the end of the previous page we derived the following expression for the RMS power delivered to a purely resistive load: $$P_{rms} = I_{rms}^2 R$$ In our example problem here we get: $$P_{rms} = (17.32)^2(12)$$ So for a 12 ohm resistor, the power is:

$$P_{rms} = (3.6 \times 10^3 ) \; W = 3.6 \; kW$$