An altering voltage is given by - e - e1 sin- t - e2 cos- t - Then the root mean square value of voltage-

The correct option is C √(e_1^2 + e_2^2) The given alternating voltage is, #10; #10; e = #10;e_1sinω t + e_2cosω t #10; #10; e = #10; ...

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Standard XII

Mathematics

Higher Order Derivatives

An altering v...

Question

An altering voltage is given by ; $e = e_{1} sin ω t + e_{2} cos ω t$ . Then the root mean square value of voltage.

\sqrt{e_{1}^{2} + e_{2}^{2}}

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\sqrt{e_{1} e_{2}}

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\sqrt{\frac{e_{1} e_{2}}{2}}

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\sqrt{\frac{e_{1}^{2} e_{2}^{2}}{2}}

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Solution

The correct option is C $\sqrt{e_{1}^{2} + e_{2}^{2}}$
The given alternating voltage is,

$e = e_{1} sin ω t + e_{2} cos ω t$

$e = e_{1} sin ω t + e_{2} sin (ω t + \frac{π}{2})$

Now, the root mean square value is,
$e = \sqrt{({e_{1}}^{2} + {e_{2}}^{2} + 2 e_{1} e_{2} cos \frac{π}{2})}$

$e = \sqrt{{e_{1}}^{2} + {e_{2}}^{2}}$

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An altering voltage is given by ; e=e1sinωt+e2cosωt . Then the root mean square value of voltage.

The correct option is C √e21+e22The given alternating voltage is, e=e1sinωt+e2cosωt e=e1sinωt+e2sin(ωt+π2) Now, the root mean square value is,e=√(e12+e22+2e1e2cosπ2) e=√e12+e22

An altering voltage is given by ; $e = e_{1} sin ω t + e_{2} cos ω t$ . Then the root mean square value of voltage.