Respuesta :
Correct Question:
A non-sinusoidal voltage is [tex]v(t) = 5 + 10sin(2\pi 50t + 30) + 15sin(4\pi 50t + 45)[/tex]This voltage is connected to the load which is a serial connection of a 10Ω resistor and a 10mH inductance. Determine: a) the power absorbed by the load, and b) Drive an expression for the load current
Answer:
a) Power absorbed by the load = 15.1 W
b) Expression for the load current: i(t) = 0.5 + 0.95sin(2π50t + 12.57) + 1.27sin(4π50t + 12.871)
Explanation:
when V₁(t) = 5V
i₁(t) = V₁(t)/R = 5/10 = 0.5 A
When V₂(t) = 10sin(2π50t + 30)
i(t) = V(t)/(R +jX)
[tex]X_{L} = 2\pi fL\\X_{L} = 2\pi * 50 * 0.01\\X_{L} = 3.14\\V(t) = 10sin(2\pi 50t + 30)[/tex]
V₂(t) = 10∠30
i₂(t) = (10∠30)/(10 + j3.14)
i₂(t) = (10∠30)/(10.481∠14.432)
i₂(t) = 0.95∠12.47 A
For source V₃(t) = 15sin(4π50t + 45) = 15∠45
i₃(t) = V₃(t)/(R + jX)
[tex]X_{L} = 2\pi FL\\X_{L} = 2\pi*100*0.01\\X_{L} = 2\pi \\R + jX_{L} = 10 + j6.284\\[/tex]
1₃(t) = (15∠45)/(10 + j6.284)
1₃(t) = (15∠45)/(11.808∠32.129)
1₃(t) = 1.27∠12.871 A
The expression for the load current becomes:
i(t) = i₁(t) + i₂(t) + i₃(t)
i(t) = 0.5 + 0.95∠12.57 + 1.27∠12.871
i(t) = 0.5 + 0.95sin(2π50t + 12.57) + 1.27sin(4π50t + 12.871)
a) The power absorbed by the load
P = I²R
I² = 0.5² + (0.95/√2)² + (1.27/√2)²
I² = 1.51 A
P = 1.51 * 10
P = 15.1 W
Answer:
We calculate for v(t) = 5 + 10sin(2π50t + 30) + 15sin(4π50t)
a) 15.1 W
b) R = √Z² - (wL)²
Explanation:
a) Let us calculate the current by considering each component of the supply voltage separately, 5/10 = 0.5 A
Using impedance RL series pair, Z = √R² + (wL)², w is different for the two component
We can obtain total current using, w = 2π50:
i = 0.5 + 10/(√100 + π²)*sin(wt + α) + 15/(√100 + 4π²)*sin(2wt + β)
Phase α and β which is not relevant in this approach, will be used later.
Then the instantaneous real power in the resistor, the P= i²R = 10i²
Let us abbreviate i = a + b + c, by squaring:
i² = a² + b² + c² + 2ab + 2ac + 2bc
Ignoring the amplitude we have:
sin(wt + α)sin(2wt + β)
By expansion, with phase r inclusive, each sine using standard trig identities gives:
{sin(wt)cos(r) + cos(wt)sin(r)}{2sin(wt)cos(wt)} = 2cos(r)sin²(wt) + 2sin(r)sin(wt)cos²(wt)
By summing averages to zero, we can ignore it. It now leaves us with just a² + b² + c² terms to average:
P = 10i = 10(0.25 + 100/(100 + π²)*sin²(wt + α) + 225/(100 + 4π²)*sin²(2wt + β)
The average of sine squared is 0.5 irrespective of the frequency and phase, hence, power,P dissipated by RL:
P = 25 + 500/(100 + π²) + 1125/(100 + 4π²) = 15.1 W
b) To derive an expression for the load, R, we have:
From the impedance, Z = √R² + (wL)²
Making R subject of formular, square both sides"
Z² = R² + (wL)²
R² = Z² - (wL)²
∴ R = √Z² - (wL)²
