WBJEE · Physics · Current Electricity
Two wires of same radius having lengths \(l_{1}\)
and \(I_{2}\) and resistivities \(\rho_{1}\) and \(\rho_{2}\) are connected in series. The equivalent resistivity will be
- A \(\frac{\rho_{1} l_{2}+\rho_{2} I_{1}}{\rho_{1}+\rho_{2}}\)
- B \(\frac{\rho_{1} l_{1}+\rho_{2} /_{2}}{l_{1}+l_{2}}\)
- C \(\frac{\rho_{1} l_{1}-\rho_{2} /_{2}}{l_{1}-l_{2}}\)
- D \(\frac{\rho_{1} l_{2}+\rho_{2} l_{1}}{l_{1}+l_{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\rho_{1} l_{1}+\rho_{2} /_{2}}{l_{1}+l_{2}}\)
Step-by-step Solution
Detailed explanation
\(R=\frac{\rho l}{A}\) where, \(\rho=\) resistivity \(l=\) length of the resistance wire \(A=\) area of cross-section When the wires are connected in senes, then…
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