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TS EAMCET 9-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 108
Total: 160
The resistance in following circuit are
R
1
=
R
2
=
R
3
=
6.0
Ω
. The emf of the battery is 12 volts. When switch S is closed, the potential across resistance
R
1
is changed by an amount.
-2 V
+2 V
-4 V
+4 V
Validate
Solution:
Initially we have following circuit, when switch S is open
Total resistance of circuit
=
6
+
6
=
12
Ω
Current in circuit,
I
=
12
V
12
Ω
=
1
A
So, potential across
R
1
is
V
A
B
=
I
A
B
×
R
1
=
1
×
6
=
6
V
Now, after switch is closed we have following circuit,
Total resistance is now,
R
e
q
=
6
+
(
6
|
|
6
)
=
6
+
6
×
6
6
+
6
=
6
+
3
=
9
Ω
Current through cell,
I
=
V
R
e
q
=
12
9
=
4
3
A
Since, the value of resistance
R
1
and
R
2
are same as
6
Ω
, hence current
I
is equally divided through
R
1
and
R
2
,
∴
I
1
=
I
2
=
I
2
=
4
3
×
2
=
2
3
A
So, potential across
R
l
is
V
A
B
′
=
I
1
R
1
=
2
3
×
6
=
4
V
So, change in potential of
R
1
is
=
V
A
B
′
−
V
A
B
=
4
−
6
=
−
2
V
© examsnet.com
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