Substitute known quantities and solve for the unknown quantity. (cont.) Solving Ohm's law for the instantaneous current gives Av (165 V)sin(50st) R R and substituting known values gives Av (165 V)sin(50xt) R R (165 V)sin [50x(4.45 x 10-³ s)] 276 Ω The unknown quantity to be determined in part (e) is the instantaneous power dissipated by the resistor when t = 4.45 x 10-³ s. The instantaneous power dissipated by the resistor is given by P = 12R. What = = 0.385 A.

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Find the instantaneous power. Use image provided
"art
Substitute known quantities and solve for the unknown quantity. (cont.)
Solving Ohm's law for the instantaneous current gives
Av (165 V)sin(50xt)
R
R
and substituting known values gives
Av (165 V)sin(50xt)
R
R
(165 V)sin [50x(4.45 x 10-³ s)]
= 0.385 A.
276 Ω
The unknown quantity to be determined in part (e) is the instantaneous power dissipated by the resistor when
t = 4.45 x 10-³ s. The instantaneous power dissipated by the resistor is given by P = 1²R.
What instantaneous power is dissipated by the resistor at t = 4.45 x 10-³ s?
W
Transcribed Image Text:"art Substitute known quantities and solve for the unknown quantity. (cont.) Solving Ohm's law for the instantaneous current gives Av (165 V)sin(50xt) R R and substituting known values gives Av (165 V)sin(50xt) R R (165 V)sin [50x(4.45 x 10-³ s)] = 0.385 A. 276 Ω The unknown quantity to be determined in part (e) is the instantaneous power dissipated by the resistor when t = 4.45 x 10-³ s. The instantaneous power dissipated by the resistor is given by P = 1²R. What instantaneous power is dissipated by the resistor at t = 4.45 x 10-³ s? W
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