Find the potential difference Va-Vb between the points a and b shown in each part of the figure (31-E14).
Figure
Answers
(a) q=q1+q2 ...(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
⇒q2 = 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 = 4 C
∴Va−Vb = 2−4/2
= 2−2
= 0 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.
Explanation:
(a) The potential difference is shown as q = q1 + q2 …(i)
When Kirchhoff’s voltage law is applied in the loop, we get
q24 q22 - q14 = 0 ⇒3 q2 = q1 …(ii)
When Kirchhoff’s voltage law is applied in the loop DCBAD, we obtain
q2 + q14 – 12 = 0
So, q2 = 48 …(iii)
From the above two equations, we get q2 = 4811.
Now, Va – Vb = 1211 V
(b) The charge be q.
Now, after applying Kirchhoff’s voltage law, we obtain
q2 + q4 – 24 + 12 = 0
⇒ Va – Vb = -16 μC2 μF=-8 V(c)
Va – Vb = 2 – 2 - q2 μF
In the loop, through all branches, the flowing net charge is q = 72 μC
Net capacitance,
C = 4 + 2 + 1 = 7 μF
The total potential difference (V) between points a and b is given by
V=qC⇒V=727=10.3 VAs
The negative terminals of the batteries are connected to a, the net potential between points a and b is -10.3 V.
