V R - V C are output voltages across the resistor and capacitor respectively- Choose a correct option- - V R -A- As frequency of source increases- V C increases-B- As frequency of source increases- V C decreases-C- V R is independent of frequency-D- V R decreases with an increment in frequency-

The correct option is B As frequency of source increases, V_C decreases.V_R = iR & i = ε/z = ε/√(R^2 + x_c^2) x_c = 1/wc With an increment in ω, x_c dec ...

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Byju's Answer

Standard XII

Physics

Alternating Current Applied to an Inductor

V R & V C are...

Question

$V_{R} & V_{C}$ are output voltages across the resistor and capacitor respectively. Choose a correct option.

As frequency of source increases,

V_{C}

increases.

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As frequency of source increases,

V_{C}

decreases.

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V_{R}

is independent of frequency.

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V_{R}

decreases with an increment in frequency.

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Solution

The correct option is B As frequency of source increases, $V_{C}$ decreases.
$V_{R} = i R & i = \frac{ε}{z} = \frac{ε}{\sqrt{R^{2} + x_{c}^{2}}}$
$x_{c} = \frac{1}{w c}$
With an increment in $ω$ , $x_{c}$ decreases, so $i$ increases; hence, $V_{R}$ increases. Also $ε = \sqrt{V_{R}^{2} + V_{C}^{2}}$ , so as $V_{R}$ increases, $V_{c}$ decreases.

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VR & VC are output voltages across the resistor and capacitor respectively. Choose a correct option.

The correct option is B As frequency of source increases, VC decreases.VR=iR & i=εz=ε√R2+x2c xc=1wc With an increment in ω, xc decreases, so i increases; hence, VR increases. Also ε=√V2R+V2C, so as VR increases, Vcdecreases.

$V_{R} & V_{C}$ are output voltages across the resistor and capacitor respectively. Choose a correct option.