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An alternating e.m.f. of peak value 350 ...

An alternating e.m.f. of peak value 350 V is applied across an ammeter of resistance 100 ohm. What will be the reading of ammeter ?

Text Solution

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Here, `E_(0) = 350 V, R = 100 Omega, I_(v) = ?`
`I_(v) = (I_(0))/(sqrt 2) . (E_(0))/(R sqrt 2) = (350)/(100 xx 1.414) = 2.47 A`
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