Question

An alterating current through a resistor of resistance 5 ohm is
I = 7sin wt + √15 coswt. the peak
value of voltage across the resistor is :-​

Answer 1

Answer:

try to solve like it

Explanation:

Question

An alternating voltage is given by e = (6sinω t + 8cosω t) volt . The peak value of voltage is given by:

Answer · 0 votes

e = 6sinω t + 8cosω t e = 6cos(ω t - 90) + 8cosω t Taking maximum values e = 6 -90 + 8 0 e = 6cos 90 - 6jsin 90 + 8 e = 8 - 6j e = 10 -36.86 e = 10cos (ω t - 36.86) So, maximum or peak value = 10

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Answer 2

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