I - 6cos wt - 8sin wt is applied across a resistor of 40 - Find the potential difference across the resistor-

The correct option is C 400 VHere I=10(610cosω t+810sinω t) =10cos(δ ω t) |I|=10A V=IR=400V ...

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Byju's Answer

Standard XII

Physics

AC Applied to a Resistor

I = 6cos wt +...

Question

$I = 6 cos w t + 8 sin w t$ is applied across a resistor of 40 $Ω$ . Find the potential difference across the resistor.

660 V

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80 V

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330 V

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400 V

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Solution

The correct option is C 400 V
Here $I = 10 (\frac{6}{10} cos ω t + \frac{8}{10} sin ω t)$

$= 10 c o s (δ - ω t)$

$| I | = 10 A$

$V = I R = 400 V$

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I=6coswt+8sinwt is applied across a resistor of 40 Ω. Find the potential difference across the resistor.

The correct option is C 400 VHere I=10(610cosωt+810sinωt)=10cos(δ−ωt)|I|=10AV=IR=400V

$I = 6 cos w t + 8 sin w t$ is applied across a resistor of 40 $Ω$ . Find the potential difference across the resistor.

The correct option is C 400 V
Here $I = 10 (\frac{6}{10} cos ω t + \frac{8}{10} sin ω t)$

$= 10 c o s (δ - ω t)$

$| I | = 10 A$

$V = I R = 400 V$