Answer this question. taken from harish chandra Verma's book.
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Answers
Answer:
(a) q=q1+q2
V
...(i)
On applying Kirchhoff's voltage law in the
loop CabDC, we get
(q2/2)+(q2/4)-(q1/4) = 0
→ 2q2+q2-q1=0
→ 3q2=q1
...(ii)
On applying Kirchhoff's voltage law in the loop DCBAD, we get
(q/2)+(q1/4)-12 = 0
→ (q1+q2)/2+(q1/4)-12 = 0
→ 3q1+2 q2 = 48
...(iii)
From eqs. (ii) and (iii), we get
9q2+2q2 = 48
11q2 = 48
→ 92 = 48/11
Now, Va-Vb = q2/4 μF
= 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
3q/4 = 12
→q = 16 μC
Now, Va-Vb = -q/2μF
→ Va-Vb = -16 μC/2 µF
= -8 V
(c) Va-Vb-2-(2-q)/2 μF
In the loop, 2+2-q/2-q/2 = 0
→q=4C
Va-Vb = 2-4/2
= 2-2
= OV
V
(d) Net charge flowing through all branches,
q=24+24+ 24 = 72 µC
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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Explanation:
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(a)
Using Kirchhoff voltage loop,
−12+2μF2Q+2μFQ1+4μFQ1=0------- (1)
In close circuit PQRSP,
−12+2μFQ+4μFQ+Q1=0 --------- (2)
From equn (1) and (2),
=>2Q+3Q1=48--------- (3)
& 3Q−Q1=48.
Substituting Q=4Q1
(3)=> 2Q+3Q1=48
8Q1+3Q148
11Q1=48Q1=1148
Vab=4μFQ1=11×448=11
(a)
Using Kirchhoff voltage loop,
−12+2μF2Q+2μFQ1+4μFQ1=0------- (1)
In close circuit PQRSP,
−12+2μFQ+4μFQ+Q1=0 --------- (2)
From equn (1) and (2),
=>2Q+3Q1=48--------- (3)
& 3Q−Q1=48.
Substituting Q=4Q1
(3)=> 2Q+3Q1=48
8Q1+3Q148
11Q1=48Q1=1148
Vab=4μFQ1=11×448=11