KVL

14 Answers

24
eureka123 ·

Circuit is short abt CD
now u know hte answer..[1]

3
iitimcomin ·

5 meu C

24
eureka123 ·

how did u get that ???

3
iitimcomin ·

current in left and right circuits are 1A .........

eure it is shorted because the capacitor is full and it dosnt allow current to pass through it so potential in the left is -1V and in the right is -2V

(-1) - (-2) = 6+q/C or q = -5μC

SRY EDITED .. SILLY MISTAKE

-2 - (-1) = 6 + q/c

-7 = q/c

q = -7 μC

24
eureka123 ·

hey sharman..part Cd is shorted becz Vc=Vd ...its connected by conducting wire....

and i dont think ans is 5 either.....asish told tha tans was 7 maybe..i dont remember exactly..

3
iitimcomin ·

cd is shortyed because its in steady state dude ...... capacitor is fully charged so acts like infinite ressistance ...

ans will be seven if the battery terminals are flipped...

3
iitimcomin ·

oh!!!! shit ...... ive made a mistake....

the potential in the left is -2 and that in the right is -1

so we have

-2 - (-1) = 6 + q/c

-7 = q/c

q = -7 μC

1
shivendra pathak ·

at steady state;current through branch CD is 0.
applying kirchoffs 2.nd law in the loop CEDC;
6-q/1=0
so; q=6 uc.

106
Asish Mahapatra ·

6/7 ?? im confused

24
eureka123 ·

i think sharman ahs made a mistake....

1
RAY ·

eureka seems right..the Vc = Vd so no current will flow through the capacitor..so no charge :))...so it will have zero resistance [not infinite becoz for getting charge the current has to pass]

3
msp ·

hey eure and rohit one thing u need to understand,if current is zero then it doesnt mean charge on the capacitor is zero.

dq/dt=0;q=c; okie.

dont get confused by taking c=0

1
RAY ·

but initially the capacitor is considered at zero charge isntit MSp??

1
RAY ·

we integrate it from 0 tO Q with respect to time???n its not mentioned that the capacitor is initially charged..so cant we make that assumption???

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