(B) When two wires are connected in series,the total resistance $R$ is the sum of individual resistances $R_1$ and $R_2$.$R = R_1 + R_2$Using the form

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(B) When two wires are connected in series,the total resistance $R$ is the sum of individual resistances $R_1$ and $R_2$.$R = R_1 + R_2$Using the formula $R = \frac{\rho L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area:$\frac{\rho(2l)}{A} = \frac{\rho_1 l}{A} + \frac{\rho_2 l}{A}$Dividing both sides by $\frac{l}{A}$:$2\rho = \rho_1 + \rho_2$$\therefore \rho = \frac{\rho_1 + \rho_2}{2}$

Class 12 Physics Current Electricity Resistance of wire, Resistivity and Conductivity

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Two wires of equal length and equal diameter and having resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is . . . . . . .

GUJCET 2015 All GUJCET

A
$\rho_1+\rho_2$
B
$\frac{\rho_1+\rho_2}{2}$
C
$\frac{\rho_1 \rho_2}{\rho_1+\rho_2}$
D
$\sqrt{\rho_1 \rho_2}$

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Current Electricity

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Resistance of wire, Resistivity and Conductivity

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Two wires of equal length and equal diameter and having resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is . . . . . . .

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$A$ nichrome wire $50 \text{ cm}$ long and $1 \text{ mm}^2$ cross-section carries a current of $4 \text{ A}$ when connected to a $2 \text{ V}$ battery. The resistivity of the nichrome wire in $\Omega \cdot \text{m}$ is:

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If copper and silicon are cooled from $300 \, K$ to $60 \, K$,their resistivity will:

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