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+1 vote
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in Physics by (60.8k points)
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Find the potential difference Va - Vb between the points a and b shown in each part of the figure (31-E14).

2 Answers

+1 vote
by (15.0k points)
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Best answer

(a)

Nodal analysis at x

2(x - 12) + 4(x - 0) + 2(x - y) = 0

4x - y = 12  ......(1)

Nodal analysis at y

4(y - 0) + 2(y - x) = 0

3y = x     ......(2)

Putting in (1)

4(3y) - y = 12

y = \(\frac {12}{11}\) volt

So,

\(V_a - V_b = \frac{12}{11} - 0\)

\(V_a - V_b = \frac{12}{11} \) volt

(b)

Emfeq = 24 - 12 = 12V

Ceq = \(\frac{2\times 4}{2 + 4} = \frac 43 \) µF

Q = CeqV

Q = \(\frac 43 \times 12\) = 16 µC

So,

\(V_a - V_b = \frac QC\)

\(= \frac{-16\mu }{2\mu }\)

\(= - 8 \,volt\)

(c)

Nodal Analysis at X

2(x - 4) + (x - 2)C + (x - 0)2 = 0

4x + Cx = 8 + 2C

x(4 + C) = 2(4 + C)

x = 2volt

So,

Va - Vb = 2 - 2 = 0 volt

(d)

Nodal analysis at Vb

(Vb - 6)4 + (Vb - 12)2 + (Vb - 24) = 0

7Vb = 24 + 24 + 24

Vb = \(\frac{72}7\)

So,

Va - Vb = 0 - \(\frac{72}7\)

Va - Vb = -10.3 volt

+4 votes
by (150k points)

(a) By loop method application in the closed circuit ABCabDA

From the figure it is cleared that the left and right branch are symmetry and reversed, so the current go towards BE from BAFEB same as the current from EDCBE.
∴ The net charge Q = 0 ∴ V = Q/C = 0/C = 0 ∴ Vab = 0
∴ The potential at K is zero.

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