An AC voltage source V - 10 sint volts is applied to the following network- Assume that R1 - 3 k- R2 - 6 k- and R3 - 9 k- and that the diode is ideal-RMS current Irms in mA through the diode is -

The equivalent resistance across terminal at (outer loop) is nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp ...

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Circuit Theory

Equivalent Representations

An AC voltage...

Question

An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that $R_{1}$ = $3 k Ω$ , $R_{2}$ = $6 k Ω$ , and $R_{3}$ = $9 k Ω$ , and that the diode is ideal.

RMS current $I_{r m s} (i n m A)$ through the diode is _
__.

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Solution

The correct option is A 1

The equivalent resistance across terminal at (outer loop) is

10 sin t
$V = \frac{1}{3} \times 3 k Ω + \frac{1}{6} \times 6 k Ω + \frac{1}{3} \times 9 k Ω$

$V = 5 I$

$o r \frac{V}{I} = 5 k Ω$

For half wave rectifier

$I_{r m s} = \frac{I_{m}}{(2)} = \frac{10 sin t}{5 k Ω} = 2 sin t m A$

$∴ I_{r m s} = \frac{I_{m}}{2} = 1 m A$

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An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that R1 = 3 kΩ, R2 = 6 kΩ, and R3 = 9 kΩ, and that the diode is ideal. RMS current Irms (in mA) through the diode is ___ ____.

The correct option is A 1 The equivalent resistance across terminal at (outer loop) is 10 sin t V=13×3kΩ+16×6kΩ+13×9kΩ V=5I orVI=5kΩ For half wave rectifier Irms=Im(2)=10sint5kΩ=2sintmA ∴Irms=Im2=1mA

An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that $R_{1}$ = $3 k Ω$ , $R_{2}$ = $6 k Ω$ , and $R_{3}$ = $9 k Ω$ , and that the diode is ideal.

RMS current $I_{r m s} (i n m A)$ through the diode is _
__.