Determine I as a function of resistance x in the circuit
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Explanation:
The given circuit is equivalent to the balanced Wheatstone's network.
Let, resistance in arm AB is P,P=15+6=21Ω
Resistance in arm BC is Q,Q=(
8+X
8X
)+(3)Ω
Resistance in arm DA is R,R=(
6+6
(6)(6)
)+(15)=3+15=18Ω
Resistance in arm CD is S,S=(
4+4
(4)(4)
)+(4)=2+4=6Ω
Using balancing condition for Wheatstone's network.
Q
P
=
S
R
(
8+X
8X
)+(3)
21
=
6
18
(
8+X
8X
)+(3)
21
=3
(
8+X
8X
)+(3)=
3
21
(
8+X
8X
)+(3)=7
8+X
8X
=4
8X=32+4X
4X=32
X=8Ω
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