Answer to Question #167254 in Trigonometry for Vishal

Question

The value of the voltage in an ac circuit at any time t seconds is given by V1 = 70 sin (50 t). Going through a component results in the wave lagging by 0.32 radians, the result being V2 = 70 sin (50π t – 0.32).

a) For t = 0 to t = 80ms, use increments of 5ms plot the following 3 graphs on the same pair of axes.

V1 = 70 sin (50 t).

V2 = 70 sin (50π t – 0.32).

V3 = V1 + V2

b) For V2 = 70 sin (50π t – 0.32) state the voltage at t = 0, 20 and 50ms.

c)Verify the answers to (b) using sin (A – B) = sin A cos B – cos A sin B.

d)Use the sine and cosine rules to find the equation for V3.

Answer 1

a. The plot is as shown below

b. $V_2 = 70 sin (50π t – 0.32)$

At t=0, $V_2 = 70 sin (50π *0 – 0.32)=70\sin \left(50\pi 0-\frac{8}{25}\right)$

$=70\sin \left(-\frac{8}{25}\right)=70\left(-\sin \left(\frac{8}{25}\right)\right)=-70\sin \left(\frac{8}{25}\right)=-22.02$

At t=20, $V_2 = 70 sin (50π *20 – 0.32)=70\sin \left(50\pi *20-\frac{8}{25}\right)$

$70\sin \left(1000\pi -\frac{8}{25}\right)=-22.02$

At t=50, $V_2 = 70 sin (50π *50 – 0.32)=70\sin \left(50\pi *50-\frac{8}{25}\right)$

$70\sin \left(2500\pi -\frac{8}{25}\right)=-22.02$

c. $sin (A – B) = sin A cos B – cos A sin B.$

$A=50πt \space and \space B=0.32$

At t=0, $sin(50\pi*0)cos0.32–cos(50\pi*0)sin0.32.=-22.02$

At t=20, $sin(50\pi*20)cos0.32–cos(50\pi*20)sin0.32.=-22.02$

At t=50, $sin(50\pi*50)cos0.32–cos(50\pi*50)sin0.32.=-22.02$

d. $V_1+V_2=\left\{70\:sin\:\left(50\pi \:t-\:0.32\right)\right\}+\left\{70\:sin\:\left(50\pi \:t\right)\right\}$

$V_3=70(\sin \left(50\pi t\right)\cos \left(0.32\right)+\cos \left(50\pi t\right)\sin \left(0.32\right))$

$V_3=70\left(\sin \left(50\pi t\right)+0.94923 \sin \left(157.07963t\right)+0.31456\cos \left(157.07963 t\right)\right)$

$V_3=70\left(1.94923\sin \left(157.0796t\right)+0.31456\cos \left(157.07963t\right)\right)$