Question

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:

• A. $40volt$
• B. $60volt$
• C. $80volt$
• D. $120volt$

Answer 1

The correct option is C $80volt$

Given, $c_1=1\mu f,c_2=2\mu f,PD=120v$

$C_{eq}=\frac{c_1{c}_2}{c_1+c_2}=\frac{2\times 1}{2+1}=\frac{2}{3}\mu f$

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

$Q_{net}=12\times \frac{2}{3}=80$

$Q=cv=1\mu fv=80\Rightarrow v=80v$