Question

A 1muF capacitor and a 2µF capacitor are connected in series across a 1200 V supply line. a. Find the charge on each capacitor and the voltage across them. b. The charged capacitors are disconnected from the line and from each other and reconnected with terminals of like sign together. Find the final charge on each and the voltage across them.

Answer 1

Given:

Capacitance $(C_1) = 1 \mu F$

                       $C_2 = 2 \mu F$

Voltage$(V) = 1200 V$

To find:

(a) The charge on each capacitor and the voltage across them.

(b) The final charge on each and the voltage across them when the charge capacitors are disconnected.

Solution:

(a) The charge on each capacitor and the voltage across

When capacitors are connected in series net capacitance is given by

                      $C_(net) = \frac{C_1C_2}{C_1+C_2}$

                                  $=\frac{2}{3} \mu F$  

We know that,

                  $q_(net) = C_(net) V$

                             $=(\frac{2}{3} \mu F)(1200V)$

                           $q_(net)=800\mu C$

In series combination q remains same.

∴                            $q_1=q_2=800 \mu C$

                                  $V_1 = \frac{q_1}{C_1} \\ = \frac{800}{1} \\ =800V$

and                           $V_2 = \frac{q_2}{C_2} \\ = \frac{800}{2} \\ =400V$

Thus,                $V_1= 800V$

                         $V_2= 400V$          

b. Now, total charge will become 1600μC. This will now distribute in direct ratio of capacity.

               ∴          $\frac{q_1}{q_2} = \frac{C_1}{C_2} = \frac{1}{2}$

                  $q_1= \frac{1}{3} (1600)= \frac{1600}{3} \mu C$

                     $q_2= \frac{2}{3} (1600)= \frac{3200}{3} \mu C$

They will have a common potential (in parallel) given by

               $V= \frac{Total charge}{Total capacity}$

                     =$\frac{1600 \mu C}{ 3 \mu F}$

                  $V =\frac{1600}{3} V$

Thus, Voltage across both capacitors is $\frac{1600}{3} V$