In the circuit shown E,F,G and H are cells of e.m.f. 2V,1V,3V and 1V respectively and their internal resistances are 2Ω,1Ω,3Ω and 1Ω respectively.
A
VD−VB=213V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
VD−VB=−213V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
VG=2113V= potential difference across G
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
VH=1913V= potential difference across H
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is DVH=1913V= 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 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=−213V
Backsubstituting the value of x, i1=513A,i2=113A,i3=613A
Potential difference across G VG=3−3i3=3−3613=2113V