NEET-XII-Physics
31: Capacitors
- #19A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?Ans : Given:
`` {C}_{1}=100\,\mathrm{\,pF\,}``
`` V=24\,\mathrm{\,V\,}``
Charge on the given capacitor, `` q={C}_{1}V=24\times 100\,\mathrm{\,pC\,}``
Capacitance of the uncharged capacitor,`` {C}_{2}=20\,\mathrm{\,pF\,}``
When the charged capacitor is connected with the uncharged capacitor, the net charge on the system of the capacitors becomes
`` {q}_{1}+{q}_{2}=24\times 100\,\mathrm{\,qC\,}...\left(\,\mathrm{\,i\,}\right)``
The potential difference across the plates of the capacitors will be the same.
Thus,
`` \frac{{q}_{1}}{{C}_{1}}=\frac{{q}_{2}}{{C}_{2}}``
`` \Rightarrow \frac{{q}_{1}}{100}=\frac{{q}_{2}}{20}``
`` \Rightarrow {q}_{1}=5{q}_{2}...\left(\,\mathrm{\,ii\,}\right)``
From eqs. (i) and (ii), we get
`` {q}_{1}+\frac{{q}_{1}}{5}=24\times 100\,\mathrm{\,pC\,}``
`` \Rightarrow 6{q}_{1}=5\times 24\times 100\,\mathrm{\,pC\,}``
`` \Rightarrow {q}_{1}=\frac{5\times 24\times 100}{6}\,\mathrm{\,pC\,}``
`` \,\mathrm{\,Now\,},``
`` {V}_{1}=\frac{{q}_{1}}{{C}_{1}}``
`` =\frac{5\times 24\times 100\,\mathrm{\,pC\,}}{6\times 100\,\mathrm{\,pF\,}}=20\,\mathrm{\,V\,}``
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