# 12th Calculus

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suppose that at any tiem t(sec) the current i(amp) in an alternating current circuit is i=2cost+2sint. What is the peak (largest magnitude) current for this circuit?

• 12th Calculus -

d(i)/dt = -2sint + 2cost
= 0 for a max of i
solving this we get sint = cost
divide by cost
sint/cost = 1
tant = 1
t = pi/4 sec

then i = 2cos(pi/4) + 2sin(pi/4)
= √2 + √2
= 2√2 amps

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