logo
or
search
upload

A 1muF capacitor and a 2µF capacitor are connected in series across a 1200 V supply line. <br> a. Find the charge on each capacitor and the voltage across them. <br> 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.

Apne doubts clear karein ab Whatsapp par bhi. Try it now.
CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

127.5 K+ views | 6.3 K+ people like this

   
Share
Share

Answer Text

Solution :
`C_(n et)=(C_1C_2)/(C_1+C_2)=2/3muF` <br> `q_(n et)=C_(n et)V` <br> `=(2/3muF)(1200V)` <br> `=800muC` <br> In series `q` remains same. <br> `:. q_1=q_2=800muC` <br> `V_1=q_1/C_1=800V` <br> and `V_2=q_2/C_2=400V` <br> b. Now, total charge will become `1600muC`. This will now distribute in direct ratio of capacity. <br> `:.q_1/q_2=C_1/C_2=1/2` <br> `q_1=(1/3)=(1600)=1600/3muC` <br> `q_2=(2/3)(1600)=((3200)/3)muC` <br> They will have a common potential (in parallel) given by <br> `V=("Total charge")/("Total capacity")` <br> `=(1600muC)/(3muF)` <br> `=1600/3V`

Related Video

4:40

125.6 K+ Views | 6.3 K+ Likes

5:09

19.6 K+ Views | 900+ Likes

8:32

16.0 K+ Views | 700+ Likes

10.9 K+ Views | 500+ Likes

4:37

15.5 K+ Views | 700+ Likes

5:05

3.4 K+ Views | 100+ Likes

87.1 K+ Views | 4.3 K+ Likes

bottom open in app