CBSE Class 12 · 2018 · Alternating CurrentLong Answer

A device X is connected across an a.c. source of voltage

V = V_{0} sin ω t

. The current through X is

I = I_{0} sin (ω t + \frac{π}{2})

. (a) Identify the device X and write the expression for its reactance. (b) Draw graphs showing the variation of voltage and current with time over one cycle of a.c. for X. (c) How does the reactance of the device X vary with the frequency of the a.c.? Show this variation graphically. (d) Draw the phasor diagram for the device X.

Answer: X is a capacitor; reactance $X_{C} = \frac{1}{ω C}$ , which falls as $1/ f$ with frequency.

Solution

(a) The current leads the voltage by

\frac{π}{2}

, so X is a capacitor. Its reactance is

X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}

(b) Over one cycle,

V = V_{0} sin ω t

and

I = I_{0} sin (ω t + \frac{π}{2})

: the current curve is shifted a quarter cycle ahead of the voltage curve (its peak comes before the voltage peak).
(c)

X_{C} = \frac{1}{2 π f C}

is inversely proportional to frequency, so the

X_{C}

–

f

graph is a rectangular hyperbola falling as

f

increases.
(d) In the phasor diagram the current phasor

I_{0}

is drawn

9 0^{\circ}

ahead of (anticlockwise from) the voltage phasor

V_{0}

CBSE Class 12 Board · 2018 · Subjective

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CBSE Class 12 · 2018 · Alternating CurrentLong Answer

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