When an alternating current of frequency f and peak current

When an alternating current of frequency f and peak current I_0 passes through a resistance R, then the power delivered to the resistance at time t seconds is P = I^2_0 R sin^2 2 pi ft. Write an expression for the power in terms of csc^2 2 pi ft. P = I^2_0 R/(csc^2 2 pi ft) P = I^2_0 R (csc^2 2 pi ft) P = I^2_0/(1 - csc^2 2 pi ft) P = I^2_0 R(1 - csc^2 2 pi ft)

Solution

P=I_o²R(sin²2ft)

P=I_o²R(sin2ft)²

write sin2ft as 1/csc(2ft)

P=I_o²R(1/csc2ft)²

P=I_o²R/(csc²2ft)