2019-03-01_Physics_Times

Let x be the potential at the junction.The potential
of (-) ve terminal of the battery is 0V and of the (+)
ve terminal is 40 V.
The potential at the junction is

1 1 2 2 3 3

1 2 3

40 4 40 4 0 4

4 4 4

V C V C V C

x

C C C

      

 

   

80

V

3

x

The charges on the two capacitors that are in
parallel are equal

80 160

4 F(40- )

3 3

C

 

Whereas the charge on the right capacitor is

80 320

4 0

3 3

F V C

 

  

 

 

The charges on the capacitors are

The total (+)ve charge supplied by the battery is

320

3

C

After closing the switch s the right side capacitor
is short circuited and can be disconnected. The
charges on the two capacitors are

The total (+)ve charge supplied by the battery is

360 .C From the above two charge distributions

After closing the switch the charge that passes

through AB (through the battery)is

320

(320 )

3

 C = 213 C

A battery always supplies equal and opposite

charges from its two terminals so we can conserve

charge on a loop that has both terminal of a battery.

Consider an indepedent loop that has two terminals

of a battery. The total charge on the loop is zero.

Two possible independent loops are shown in the

figure.

By applying Kirchoff’s law

24 12 0

4 2

   

 

q q

V V

F F

3

12 16

4

   

q

q C

The potential difference between a and b is

16

8

2 2

  

    

a b  

q C

V V V

F F

(^22) ,
Z R XL XLL and  ^2 f
  Z R^24 2 2 2f L
s p
p s
N i
N i
 or
When the switch is closed
42.Sol:
43.Sol:
44.Sol:
45.Sol:
25
50
1 2
p
p
i
  i A
The boolean expression of this gate is
(^) Y= (A+B).(A+B)=(A+B)+ (A+B)= (A+B)
which is for NOR gate.