question_answer

A series R-C circuit is connected to an alternating voltage source. Consider two situations:                                           

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.