An alternating voltage e - 200 -2 sin 100t volt is connected to 1 - F capacitor through a-c- ammeter- The reading of ammeter is

The correct option is B 20 mA E=200√(2)sin(100t) Hence, E_rms=E_o√(2)=200V #160; #160; #160; #160; #160;and #160; #160; # ...

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Standard XII

Physics

Capacitive Reactance

An alternatin...

Question

An alternating voltage e = $200 \sqrt{2}$ sin $(100 t)$ volt is connected to $1 μ F$ capacitor through a.c. ammeter. The reading of ammeter is

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Solution

The correct option is B $20$ mA
$E = 200 \sqrt{2} sin (100 t)$

Hence, $E_{r m s} = \frac{E_{o}}{\sqrt{2}} = 200 V$ and $ω = 100 s^{- 1}$

Now, impedence of capacitive circuit, $X_{C} = \frac{1}{ω C}$

$⟹ X_{C} = \frac{1}{100 \times 10^{- 6}} = 10^{4}$

Hence, $I_{r m s} = \frac{E_{r m s}}{X_{C}} = \frac{200}{10^{4}} A$

$I_{r m s} = 20 m A$

Hence, answer is option-(D).

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An alternating voltage e = 200√2 sin (100t) volt is connected to 1μF capacitor through a.c. ammeter. The reading of ammeter is

The correct option is B 20 mAE=200√2sin(100t)Hence, Erms=Eo√2=200V and ω=100s−1Now, impedence of capacitive circuit, XC=1ωC⟹XC=1100×10−6=104Hence, Irms=ErmsXC=200104AIrms=20mAHence, answer is option-(D).

An alternating voltage e = $200 \sqrt{2}$ sin $(100 t)$ volt is connected to $1 μ F$ capacitor through a.c. ammeter. The reading of ammeter is