Question

Supply is given between A and B in the figure as shown. Ratio of potential difference across 4 microFarad and 6 microFarad is

(1) 2:3. (2) 3:2. (3) 1:2. (4) 2:1.​

Answer 1

Answer:

(4). 2/1

Explanation:

This is a wheatstone bridge .

So, basically :

The diagram that will follow will be like the one attached .

We know that potential difference between parallel connection is always same .

Now , the potential between A & B will be same .

Equivalent capacitance in top part is

= 6*3/9

= 2

Equivalent capacitance in lower part is

= 4*8/12

= 8/3

Let the charge in top part be Q & that in the lower part be Q' .

So, V in upper part = V in lower part

Or, Q/2 = Q'/(8/3)

4Q = 3Q' .....(1)

Also, Q = C*V

So, potential drop in C6 (V6)= Q/6

potential drop in C4 (V4)= Q'/4

V4/V6 = (Q'/4)/Q/6

= (4Q/3*4) / Q/6 .....(from 1)

= 2

So, V4/V6 = 2:1

Answer 2

Answer:

In this solution

• Capacitor 6 and 3 in series
• So, There is C1=6×3/6+3
• C1=2
• V1=Q1/c1
• V1=Q1/2.

Capacitor 4 and 8 are also in series

C2=4×8/4+8

C2=8/3

V2=Q2×3/8

V1=v2

Q1/2=Q2×3/8

4Q1=3Q2

• Here, v6=Q/6
• V4=4Q/12

So, v4/v6=24Q/12Q

V4/v6=2/1

Ans. 2:1(d)

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