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Question : 12
Total: 27
Show how would you join three resistors, each of resistance 9 Ω 13.5 Ω 6 Ω
OR
(a) Write Joule's law of heating.
(b) Two lamps, one rated100 W ; 220 V 60 W ; 220 V 220 V
OR
(a) Write Joule's law of heating.
(b) Two lamps, one rated
Solution:
(a) To get an equivalent resistance of 13.5 Ω
=
+
=
+
=
=
=
R P =
= 4.5 Ω
R S = R 3 + 4.5 Ω
= 9 Ω + 4.5 Ω
= 13.5 Ω
Now,
R S = R 3 + 4.5 Ω
= 9 Ω + 4.5 Ω
= 13.5 Ω
(b) To get an equivalent resistance of6 Omega
R S = R 1 + R 2
= 9 + 9
= 18 Ω
Now both the resistors are in parallel with each other so,
R P =
+
=
=
=
Ω
So,R P = 6 Ω
OR
a() According to Joule's law of heating, the heat produced in a wire is directly proportional to
(i) square of current( I 2 )
(ii) resistance of wire (R),
(iii) time (t
Thus, the heat produced in the wire by currentI t
or
H ∝ I 2 R t
H = KI 2 R t
H = I 2 R t
But K = 1 ,
(b) We know that,P = VI
⇒
I =
First lamp :
P 1 = 100 W , V = 220 volt
I 1 =
=
= 0.45 A
Second lamp :P 2 = 60 W , V = 220
I 2 =
=
= 0.27 A
So, Total current
= I 1 + I 2
= 0.45 + 0.27
= 0.72 A
So,
Now,
(b) To get an equivalent resistance of
Now both the resistors are in parallel with each other so,
So,
OR
a() According to Joule's law of heating, the heat produced in a wire is directly proportional to
(i) square of current
(ii) resistance of wire (R),
(iii) time (
Thus, the heat produced in the wire by current
or
(b) We know that,
First lamp :
Second lamp :
So, Total current
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