Solution:

Ans. (i) Identification

(ii) Identifying the curves Justification

(iii) Variation of Impedance with frequency Graph

(iv) Expression for current Phase relation

(i) The device $X$ is a capacitor

(ii) Curve $B \rightarrow$ voltage

Curve $C \rightarrow$ current

Curve $A \rightarrow$ power

Reason : The current leads the voltage in phase

by $\frac{\pi}{2}$, for a capacitor.

(iii) $X_{c}=\frac{1}{\omega C}$ or $\left(X_{c} \propto \frac{1}{\omega}\right)$

$1 / 2$

$1 / 2$

$1 / 2$ $1 / 2$ $1 / 2$

(iv)

$$ V=V_{0} \sin \omega t $$

$$ \begin{aligned} & q=V C=C V_{0} \sin \omega t \ & I=\frac{d q}{d t}=\omega C V_{0} \cos \omega t \end{aligned} $$

$$ \begin{aligned} & V=V_{0} \sin \omega t \ & I=\mathrm{I}_{0} \sin \left(\omega t+\frac{\pi}{2}\right) \end{aligned} $$

Current leads the voltage, in phase, by $\frac{\pi}{2} \quad 1 / 2$ [Note : If the student identifies the device $X$ as an Inductor but writes correct answer to parts (c) and (d) (in terms of an inductor), the student be given full marks for (only) these two parts]

[CBSE Marking Scheme 2017]

AT Q. 3. (i) An $a c$ source generating a voltage $V=V_{0} \sin$ $\omega t$ is connected to a capacitor of capacitance $C$. Find the expression of the current I flowing through it. Plot a graph of $V$ and $I$ versus $\omega t$ to show that the current is $\frac{\pi}{2}$ ahead of the voltage.

(ii) A resistor of $200 \Omega$ and a capacitor of $15 \mu \mathrm{F}$ are connected in series to a $220 \mathrm{~V}, 50 \mathrm{~Hz} a c$ source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Why the algebraic sum of these voltages is more than the source voltage? A [CBSE SQP 2016]

Ans. (i) $\quad V \in V_{0} \sin \omega t, V=\frac{Q}{C}$

(V)I

(ii)

$$ X_{C}=\frac{1}{2 \pi f C}=212.3 \Omega $$

$$ \begin{array}{rlr} \mathrm{Z} & =\sqrt{R^{2}+X_{C}^{2}}=291.5 \Omega & 1 / 2 \ I_{r m s} & =\frac{V_{r m s}}{Z}=\frac{220}{291.5}=0.755 \mathrm{~A} & 1 / 2 \ V_{R}(r m s) & =151 \mathrm{~V} & \ \mathrm{~V}_{C}(r m s) & =160.3 \mathrm{~V} & 1 / 2 \end{array} $$

Two voltages are out of phase, hence they are added vectorially.

[CBSE Marking Scheme 2016]