question_answer

A series R-C circuit is connected to an alternating voltage source. Consider two situations:         [NEET (Re) 2015]

1. When capacitor is air filled.

2. When capacitor is mica filled.

Current through resistor is i and voltage across capacitor is V then

A)  \[{{V}_{a}}<{{V}_{b}}\]                   

B)  \[{{V}_{a}}>{{V}_{b}}\]

C)  \[{{i}_{a}}>{{i}_{b}}\]         

D)       \[{{V}_{a}}={{V}_{b}}\]

Correct Answer: B

Solution :

Net reactive capacitance,

\[V={{V}_{0}}\sin \omega t\]

\[{{X}_{c}}=\frac{1}{2\pi fC}\]

So, current in circuit,

\[I=\frac{V}{Z}=\frac{V}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{2\pi fC} \right)}^{2}}}}\]

\[\Rightarrow \]   \[I=\frac{2\pi fC}{\sqrt{4{{\pi }^{2}}{{f}^{2}}{{C}^{2}}{{R}^{2}}+1}}\times V\]

Voltage drop across capacitor, \[{{V}_{c}}=I\times {{X}_{c}}\]

\[=\frac{2\pi fC}{\sqrt{4{{\pi }^{2}}{{f}^{2}}{{C}^{2}}{{R}^{2}}+1}}\times \frac{1}{2\pi fC}\]

i.e.        \[{{V}_{c}}=\frac{V}{\sqrt{4{{\pi }^{3}}{{f}^{2}}{{C}^{2}}{{R}^{2}}+1}}\]

When mica is introduced, capacitance will increase hence, voltage across capacitor get decrease.