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Question
Find the potential difference `V_a - V_b` between the points a and b shown in each part of the figure.
Solution
(a) `q = q_1 + q_2 ..........(1)`
On applying Kirchhoff's voltage law in the loop CabDC, we get
`q_2/2 + q_2/4 - q_1/4 = 0`
`⇒ 2 q_2 + q_2 - q_1 = 0`
`⇒ 3 q_2 = q_1 .............(2)`
On applying Kirchhoff's voltage law in the loop DCBAD, we get
`q/2 + q_1/4 - 12 = 0`
`⇒ (q_1+q_2)/2 + q_1/4 - 12 = 0`
`⇒ 3q_1 + 2q_2 = 48 ............(3)`
From eqs. (2) and (3), we get
`9 q_2 + 2 q_2 = 48`
`⇒ 11q_2 = 48`
`⇒ q_2 = 48/11`
Now,
`V_a - V_b = q_2/(4 "uF") = 48/44 = 12/11 V`
(b) Let the charge in the loop be q.
Now, on applying Kirchhoff's voltage law in the loop, we get
`q/2 + q/4 - 24 +12 = 0`
`⇒ (3 q)/4 = 12`
`⇒ q = 16 "uC"`
Now ,
`V_a - V_b = (-q)/(2 "uF")`
`⇒V_a - V_b = (-16 "uC")/(2 "uF") = -8V`
(c) `V_a - V_b = 2-(2-q)/(2 "uF")`
In the loop,
`2+2-q/2-q/2 = 0`
`⇒ q = 4C`
`therefore` `V_a - V_b = 2-4/2 = 2-2 = 0 V`
Net Charge flowing through all branches, `q = 24+24+24 = 72 "uC"`
Net capacitance of all branches, C = 4 + 2 + 1 = 7 μF
The Total potential difference (V) between points a and b is given by
`V= q/c`
`⇒ V = 72/7 = 10.3 "V"`
As the negative terminals of the batteries are connected to a, the net potential between points a and b is -10.3 V .
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