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Question

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

Solution
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e=6sinωt+8cosωt
e=6cos(ωt90)+8cosωt
Taking maximum values
e=690+80
e=6cos906jsin90+8
e=86j
e=1036.86
e=10cos(ωt36.86)
So, maximum or peak value =10

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