In a simple AC circuit shown on the right, R = 85 ohm, delta V = delta V_max sin
ID: 1413278 • Letter: I
Question
In a simple AC circuit shown on the right, R = 85 ohm, delta V = delta V_max sin(omega t)where delta V_max = 5.5 V, and omega = 55 rad/s. Randomized Variables R = 85 ohm delta V_max = 5.5V omega = 55 rad/s Express the current I in terms of t, R, delta V_max, omega. Express the maximum of the current, I_max, in terms of delta V_max and R. Calculate the numerical value of I_max in amps. How long (in seconds) does it take for the current I to reach I_max (and be moving in the same direction) from the previous I_max? Express the power dissipated in the circuit, P, in terms of I and delta V. Express the power dissipated in the circuit, P, as a function of time t and R, delta V_max, and omega. Express the maximum of P, P_max, in terms of R and delta V_max. Calculate the numerical value of P_max in watts. How long does it take for the power P to reach P_max from the previous P_max in seconds.Explanation / Answer
(a) ans
the current I=Iosin(wt)
but Io=Vmax/R
so the current I=[Vmax/R] sin(wt)
(b) ans
according to the ohms law=>V=IR
basing on alternating current
maximum current Imax=Vmax/R
(c) ans
maximum current Imax=Vmax/R=5.5/85=0.065A
(d) ans
I=Imax sin (wt)
I=Imax sin(55t)
here one hint is I=Imax
=>imax=Imax sin(55t)
sin^-1[1]=55t
90=55t
therefore the time to convert I is Imax=>t=90/55=1.64 sec
(e) ans
the current I=Iosin(wt)
voltage V=Vo sin(wt)
according to the joules law =>power=IV
(f) ans
power disspated in the circuit =>p=V^2/R=(Vmax sin(wt))^2/R=[Vmax^2/R]sin^2(wt)
(g) ans
power P=[Vmax Imax]/2 cos(theta)
thet means Pmax=Vmax.Imax/2=Vmax^2/2R [Imax=Vmax/R]
(h) ans
the power Pmax =Vmax^2/2R=(5.5)^2/2*85=0.18 Waits
(i) ans
the power P=Pmax cos(wt)
here we observed P=Pmax
Pmax=Pmax cos(55t)
cos^-1(1)=55t
180=55t
time t=3.3 sec