Search any question & find its solution
Question:
Answered & Verified by Expert
A series $R$ - $C$ circuit is connected to $A C$ voltage source. Consider two cases; $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true?
Options:
Solution:
1649 Upvotes
Verified Answer
The correct answers are:
$I_R^A < I_R^B$,
$V_C^A>V_C^B$
$I_R^A < I_R^B$,
$V_C^A>V_C^B$
$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}$
In case (b) capacitance $C$ will be more.
Therefore, impedance $Z$ will be less.
Hence, current will be more.
$\therefore$ Option (b) is correct.
Further, $\quad \begin{aligned} V_C & =\sqrt{V^2-V_R^2} \\ & =\sqrt{V^2-(I R)^2}\end{aligned}$
In case (b), since current $I$ is more.
Therefore, $V_C$ will be less.
$\therefore$ Option (c) is correct.
$\therefore$ Correct options are (b) and (c).
Analysis of Question
(i) Question is moderately difficult.
(ii) In my opinion problems of alternating currents are not very difficult.
(iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.
In case (b) capacitance $C$ will be more.
Therefore, impedance $Z$ will be less.
Hence, current will be more.
$\therefore$ Option (b) is correct.
Further, $\quad \begin{aligned} V_C & =\sqrt{V^2-V_R^2} \\ & =\sqrt{V^2-(I R)^2}\end{aligned}$
In case (b), since current $I$ is more.
Therefore, $V_C$ will be less.
$\therefore$ Option (c) is correct.
$\therefore$ Correct options are (b) and (c).
Analysis of Question
(i) Question is moderately difficult.
(ii) In my opinion problems of alternating currents are not very difficult.
(iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.