A) \[3\,\Omega \]
B) \[6\,\Omega \]
C) \[12\,\,\Omega \]
D) \[1.5\,\,\Omega \]
Correct Answer: D
Solution :
Resistance of original wire is \[R=\rho \frac{l}{A}\] \[\rho \], being the specific resistance of wire. When the wire is cut in two equal halves then resistance becomes \[R=\frac{\rho \,I/2}{A}=\frac{R}{2}\] Thus, the net resistance of parallel combination of two halves is given by \[{{R}_{net}}=\frac{R\times R}{R+R}=\frac{R}{2}=\frac{R}{2\times 2}=\frac{6}{4}=1.5\,\,\Omega \]
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