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Question : 20
Total: 28
(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b) In an electric circuit two resistors of1 2 Ω 6 V
OR
An electric lamp of resistance20 Ω 4 Ω 6 V
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp.
(b) In an electric circuit two resistors of
OR
An electric lamp of resistance
(a) the total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp.
Solution:
(a) It is observed that total current I is equal to the sum of separate currents.
I = I 1 + I 2 + I 3
LetR p
I =
I 1 =
, I 2 =
and I 3 =
So, now from equation (i), we have
=
+
+
and
=
+
+
Hence, ifn
=
+
+
+ . . . . . . . . +
(b) Given, Two resistors of12 Ω
∵
= 6 V
∴
+
=
+
=
R eq =
= 6 Ω
According to ohm's law V=IR
6 = I × 6
= I
I = 1 ampere.
OR
(a) Given,R 1 = 20 Ω , R 2 = 4 Ω
Since, in Series
R = R 1 + R 2
∴
R = 20 + 4
= 24 Ω
(b) Current through circuit :
V = 6 V , R = 24 Ω
According to ohm's law
V = IR
So,
I =
I =
=
= 0.25
(c) (i) Potential difference across conduction:
I =
, R = 20 Ω
V 1 = I 1
V 1 =
× 20 = 5 V
(ii) Potential difference across lamp
V 2 = IR 2
=
× 4
V 2 = 1 V
(d) Power of lamp :
P = I 2 R
= (
) 2 × 20
=
×
× 20 =
watt
or
P = 1.25 watt
Let
So, by applying ohm's law
So, now from equation (i), we have
and
Hence, if
(b) Given, Two resistors of
According to ohm's law V=IR
OR
(a) Given,
Since, in Series
(b) Current through circuit :
According to ohm's law
So,
(c) (i) Potential difference across conduction:
(ii) Potential difference across lamp
(d) Power of lamp :
or
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