1. When capacitor is air filled. |
2. When capacitor is mica filled. |
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.You need to login to perform this action.
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