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Capacitors in Series
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Two capacitors of $1\mu F$ and $2\mu F$ are connected in series and this combination is changed upto a potential difference of $120$ volt. What will be the potential difference across $1 \mu F$ capacitor:
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A
$80 volt$
💡 Explanation:
Given, $c _1=1\mu f,c _2=2\mu f,PD=120v$
$C _{eq}=\dfrac{c _1c _2}{c _1+c _2}=\dfrac{2\times1}{2+1}=\dfrac{2}{3}\mu f$
We know, $Q=cv$ Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.
Now, $Q _{net}$ in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD
$Q _{net}=12\times\dfrac{2}{3}=80$
$Q=cv=1\mu fv=80\Rightarrow v=80v$