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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 HLet the current through the three branches between B and D be i1,i2 and i3 as shown in the figure. Let VD−VB=x. Then, the voltage drop across all three paths of the circuit should equal x. −3i1+2−1=x -- (i) −2i2=x -- (ii) 4i3+1−3=x -- (iii) At point B, i3=i1+i2 (Kirchoff's current law) Substituting from (i), (ii) and (iii), x+24=1−x3+x2 ⇒x=−213 V Backsubstituting the value of x, i1=513 A,i2=113 A,i3=613 A Potential difference across G VG=3−3i3=3−3613=2113 V VH=1+1×i3=1913 V

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