7. Find the current in DE arm of the circuit shown in fig.
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Answer:
Given,
AB & CE is parallel to battery,
So, AB & CE have equal potential difference V
AB
=V
CE
=1.5V
Potential in AB,V
AB
=1.5V
In CE, two resisters 6Ω&4Ω in series R
total
=6+4=10Ω
Total current I
AB
=
R
AB
V
AB
=
12
1.5
=0.125A
Total current I
CE
=
R
total
V
CE
=
10
1.5V
=0.15A=I
AC
Potential V
CD
=IR
CD
=0.15×6=0.9V
Potential V
DE
=IR
DE
=0.15×4=0.6V
V
DE
=0.6V,V
CD
=0.9V&V
AB
=1.5V
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