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Question

In the circuit shown E,F,G and H are cells of e.m.f. 2 V,1 V,3 V and 1 V respectively and their internal resistances are 2 Ω,1 Ω,3 Ω and 1 Ω respectively.


A
VDVB=213 V
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B
VDVB=213 V
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C
VG=2113 V= potential difference across G
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D
VH=1913 V= potential difference across H
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Solution

The correct option is D VH=1913 V= potential difference across H
Let the current through the three branches between B and D be i1,i2 and i3 as shown in the figure.


Let VDVB=x.

Then, the voltage drop across all three paths of the circuit should equal x.

3i1+21=x -- (i)
2i2=x -- (ii)
4i3+13=x -- (iii)

At point B,
i3=i1+i2 (Kirchoff's current law)

Substituting from (i), (ii) and (iii),

x+24=1x3+x2

x=213 V

Backsubstituting the value of x,
i1=513 A,i2=113 A,i3=613 A

Potential difference across G
VG=33i3=33613=2113 V

VH=1+1×i3=1913 V

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