Sorry for X-Posting but needing help quickly, hopefully someone can assist.
The problem:
Two voltages v1 and v2 are to be summed using an analogue circuit.
v1 = 5 cos (5 pi t + (pi / 6))
v2 = 10 sin (5 pi t - (pi / 6))
Calculate the time at which the sum is first equal to 4 volts.
I have added the two voltages and left the result in R = sin (wt + a) and
was calculated in radians
w = omega a = alpha
My answer to the sum is 11.18 sin (5 pi t + 1.63)
Now, how do I find the time when the voltage is first equal to 4 volts.
Appreciate any help.
TIA

Solve the equation 11.18 sin (5 pi t + 1.63) = 4. There are an infinite
number of solutions, but you can get the smallest positive one using
the inverse sine function.
-- Christopher Heckman

Huh, you cannot finish? Solve
11.18 sin(5pi.t + 1.63) = 4
for t. It will have a number of values. I'd consider the smallest
positive value for t, to be when the voltage is first 4 volts.

Your value of 11.18 is incorrect by my reconning. I make it v = 6.196 sin(5pi t
- 0.1077)
From: v1 = 4.33cos(5pi t) - 2.5sin(5pi t)
and v2 = 8.66sin(5pi t) - 5cos(5pi t)
v1 +v2 = v = 6.16sin(5pi t) - 0.67cos(5pi t)
which gives: v = 6.196sin(5pi t - 0.1077)
when v = 4Volts we have 4/6.196 = sin(5pi t - 0.1077) = 0.6456
therefore: arcsin(0.6456) = 5pi t - 0.1077 = 0.702
solving for t gives 0.0515 seconds
If you ave access to Matlab (or similar) you can verify my calculations for
yourself.
Hope this helps
John Ferguson
Windsor, England

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