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Two wires of equal diameters of resistivities \mathrm{\rho_1\: and\: \rho_2}  and lengths \mathrm{l_1\: and\: l_2}respectively are joined in series. The equivalent resistivity of the combination is:

Option: 1

\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}
 


Option: 2

\frac{\rho_1 l_2-\rho_2 l_1}{l_1-l_2}
 


Option: 3

\frac{\rho_1 l_2+\rho_2 l_1}{l_1+l_2}
 


Option: 4

\frac{\rho_1 l_1-\rho_2 l_2}{l_1-l_2}


Answers (1)

best_answer

Resistance of a wire, \mathrm{R}=\frac{4 \rho l}{\pi \mathrm{D}^2}

Where \mathrm{l} is the length, \mathrm{\mathrm{D}} is the diameter and \mathrm{\rho} is the resistivity of the material of the wire.

As the wires are connected in series, then

\frac{4 \rho_{\mathrm{S}}\left(l_1+l_2\right)}{\pi \mathrm{D}^2}=\frac{4 \rho_1 l_1}{\pi \mathrm{D}^2}+\frac{4 \rho_2 l_2}{\pi \mathrm{D}^2} [ \because Since the wires of same diameter]

Where \rho_{\mathrm{S}} is the equivalent resistivity

\rho_{\mathrm{S}}\left(l_1+l_2\right)=\rho_1 l_1+\rho_2 l_2 \quad$ or $\quad \rho_{\mathrm{S}}=\frac{\rho_1 l_1+\rho_2 l_2}{l_1+l_2}

Hence Option 1 is correct

Posted by

mansi

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