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| 1. |
Prove that the total current which is the sum of conduction current and displacement current is always continuous and any loss in conduction current (I_C) appears as displacement current (I_D). |
| Answer» <html><body><p></p>Solution :Consider a volume V in a medium through which <br/> currents are flowing. Let `I_c` be the conduction <br/> current entering the volume V and `I_c'` be the <br/> conduction current leaving the volume V. Then <br/> total charges entering and leaving the volume in <br/> time dt will be `I_c` and `I_c'`dt. Therefore, the <br/> <a href="https://interviewquestions.tuteehub.com/tag/charge-914384" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGE">CHARGE</a> <a href="https://interviewquestions.tuteehub.com/tag/accumulated-366703" style="font-weight:bold;" target="_blank" title="Click to know more about ACCUMULATED">ACCUMULATED</a> inside the volume V during <br/> time dt is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by dq(inside V)=`I_cdt-I_c'dt` <br/> or `(dq)/(dt)=I_c-I_c'....(i)` <br/> From Gauss's Theoram in electrostatics, we have <br/> `phi_E=ointvecE.vec(<a href="https://interviewquestions.tuteehub.com/tag/ds-960390" style="font-weight:bold;" target="_blank" title="Click to know more about DS">DS</a>)=q/(in_0) ("inside") or in_0 phi_E=q` <br/> or `I_d=in_0 (dphi_E)/(dt)=(d (in_0phi_E))/(dt)=(dq)/(dt) ......(ii)` <br/> From (i) and (ii), <br/> `I_c-I_c'=I_d or I_c=I_d+I_c'` <br/> Thus we conclude that the loss of conduction <br/> current `(=I_c-I_c')` appears as the <a href="https://interviewquestions.tuteehub.com/tag/displacement-956081" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACEMENT">DISPLACEMENT</a> <br/> current `(I_d)` and conduction current plus <br/> displacement current remains constant i.e., <br/> `I_c'+I_d`= a constant.</body></html> | |