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KVPY SA Exam 2015 Question Paper
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© examsnet.com
Question : 26
Total: 80
A light bulb of resistance R = 16Ω is attached in series with an infinite resistor network with identical resistances r as shown below. A 10 V battery drives current in the circuit. What should be the value of r such that the bulb dissipates about 1 W of power.
14.8Ω
29.6Ω
7.4Ω
3.7Ω
Validate
Solution:
Let assume
R
eq
=
P
Q
=
x
R
eqPQ
=
r
+
r
x
r
+
x
x
=
r
2
+
r
x
+
r
x
r
+
x
r
x
+
x
2
=
r
2
+
2
r
x
x
2
−
r
x
−
r
2
=
0
x
=
+
r
±
√
r
2
+
4
r
2
2
⇒
r
(
1
+
√
5
)
2
Power in bulb = 1 watt
i
2
R
=
1
i
2
×
16
=
1
i
−
1
4
amp.
i
=
10
R
+
R
PQ
1
4
=
10
16
+
r
2
(
1
+
√
5
)
16
+
r
2
(
1
+
√
5
)
=
40
r
=
14.8
Ω
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