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\[R\frac{\rho l/2}{A}=\frac{R}{2}\] Thus, the net resistance of parallel combination of two halves is given by \[{{R}_{net}}=\frac{R\times R}{R+R}=\frac{R}{2}=\frac{R}{2\times 2}=\frac{6}{4}=1.5\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
