two wires of same length and area ,made of two materials of resistivity rho1 and rho2 are connected in parallel V to a source of potential. the equivalent resistivity for the combination is......
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Step-by-step explanation:
length and area ,made of two materials of resistivity rho1 and rho2 are connected in parallel V to a source of potential. the equivalent resistivity for the combination
Step-by-step explanation:
Two wires of same length and area made of two materials of resistivity \rho_1\ and\ \rho_2ρ
1
and ρ
2
. The equivalent resistance of resistors in series combination is given by :
R_{eq}=R_1+R_2R
eq
=R
1
+R
2
Since,
R=\rho\dfrac{l}{A}R=ρ
A
l
...........(1)
Where
length and area of two materials are same
\dfrac{\rho_{eq}l}{A}=\dfrac{\rho_1l}{A}+\dfrac{\rho_2l}{A}
A
ρ
eq
l
=
A
ρ
1
l
+
A
ρ
2
l
So,
\rho_{eq}=\rho_1+\rho_2ρ
eq
=ρ
1
+ρ
2
The equivalent resistance of resistors in parallel combination is given by :
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
R
eq
1
=
R
1
1
+
R
2
1
Using equation (1) we get :
\dfrac{A}{\rho_{eq} l}=\dfrac{A}{\rho_1 l}+\dfrac{A}{\rho_2 l}
ρ
eq
l
A
=
ρ
1
l
A
+
ρ
2
l
A
So,
\dfrac{1}{\rho_{eq}}=\dfrac{1}{\rho_1}+\dfrac{1}{\rho_2}
ρ
eq
1
=
ρ
1
1
+
ρ
2
1
Hence, this is the required solution.
Learn more,
Resistivity
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