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A parallel plate capacitor of plate area A each and separation d, is being charged by an ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.

Answers (1)

When the capacitor is getting charged, we have electric flux $=\phi _{E}(t)$

$=\frac{Q(t)}{\varepsilon _{o}}\; \; \; \; ---(i)$

We know that, $Q(t)=CV(t)\; \; \; \; \; \; \; \; \because (V=V_{o}\sin wt)$

$=C(V_{o}\sin wt)\; \; \; \; \; \; ----(ii)$

The displacement current

$(i_{d})=\varepsilon _{o}\frac{d\phi _{E}}{dt}$

from equation (i) and (ii), we have

$\phi _{E}=\frac{CV_{o}\sin wt}{\varepsilon _{o}}$

Now, $i _{d}=\frac{cd(V_{o}\sin wt)}{dt}$

Where, $V _{o}$ is voltage due to alternating current.

$\therefore i_{d}=CV_{o}\cos wt.w$

$i_{d}=CV_{o}\cos wt.w$

and, we know that

$\cos wt=\sin (wt+\frac{\pi }{2})$

$i_d = cw\; V_{o}\sin (wt+\frac{\pi }{2})$

also the conduction current $(i_{c})$ leads by $\pi /2\; then,$

$i_{c}=\frac{V_{o}}{X_{c}}\sin (wt+\pi /2)$ where,

$X_{c}=$ $=\frac{1}{wc}$

$i_{c}=\frac{V_{o}}{1/wc}\sin (wt+\pi /2)$

$i_{c}=wC\: V_{o} \sin (wt+\pi /2)$

hence, $i_{d}=i_{c}$

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Safeer PP

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