The terminals of a battery with 18 V is connected to a circular conducting loop of resistance 24 92 at two points A & C (as shown in fig). The current flowing through bigger part ABC is. 1) 0.5 A 2) 0.75 A 3) 1.0 A 4) 1.5 A ww
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Since the terminal is connected at one quarter distance .
The ratios of length of two points is
L
2
L
1
=
π/2
3π/2
R=24Ω
Since R∝L..so
R
2
R
1
=
L
1
L
1
=3
R
1
=18ΩandR
2
=6Ω
Since the potential across R
1
and R
2
is same ,so they both are parallel.
R
c
1
=
R
1
1
+
R
2
1
; R
c
=4.5Ω
Now the internal Resistance is 1.5Ω
R
T
=1.5+4.5=6Ω
Cuurent, I=
R
T
V
=
6
18
=3A
Now the current in parallel resistance is distributed in inverse ratio of resistance .
Curren through bigger arc R
1
I
1
=
R
1
+R
2
R
2
I=
24
6
×3=0.75A