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A wire of 3Ω resistance and 15cm in length is stretched to 45 cm length. Calculate (i) New resistance.Assuming the wire has uniform cross section area. |
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Answer» Given: Resistance R = 3Ω Original length L1 = 15cm = 0.15m New length, L2 = 45 cm = 0.45 m Formula Used: Resistance, R = \(\frac{ρL}{A}\) Where, ρ is resistivity of the wire L is the length of the wire A is the area of the wire. Now, \(\frac{R_1}{R_2}=\frac{ρL_1\,\times\,A}{A\,\timesρL_2}\) ⇒ \(\frac{3\,Ω}{R_2}=\frac{0.15m}{0.45m}\) ⇒ R2 = \(\frac{3Ω\,\times\,0.45\,m}{0.15\,m}\) = R2 = 9 Ω Hence, the new resistance is 9 Ω |
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