A) \[R\]
B) \[4R\]
C) \[\frac{R}{4}\]
D) \[\frac{R}{16}\]
Correct Answer: D
Solution :
Resistance\[R=\frac{\rho l}{A}\] when wire cut into 4 equal parts, then \[R'=\frac{\rho l}{A\cdot 4}=\frac{R}{4}\] Now, from parallel combination of resistor, \[\frac{1}{{{R}_{e{{q}^{u}}}}}=\frac{1}{\left( \frac{R}{4} \right)}+\frac{1}{\left( \frac{R}{4} \right)}+\frac{1}{\left( \frac{R}{4} \right)}+\frac{1}{\left( \frac{R}{4} \right)}=\frac{16}{R}\] \[\therefore \] \[{{R}_{e{{q}^{u}}}}=\frac{R}{16}.\]
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