A) 32 \[\Omega \]
B) 40 \[\Omega \]
C) 44 \[\Omega \]
D) 48\[\Omega \]
Correct Answer: A
Solution :
Given. 1 = 4 m, R = potentiometer wire resistance = \[8\Omega \] Potential gradient = \[\frac{dV}{dr}=1mV/cm\] So, for \[400cm,\Delta V=400\times 1\times {{10}^{-3}}=0.4V\] Let a resistor \[{{R}_{s}}\] connected in series, so as \[\Delta V=\frac{V}{R+{{R}_{s}}}\times R\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,0.4=\frac{2}{8+R}\times 8\Rightarrow 8+R=\frac{16}{0.4}=40\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,R=32\,\Omega \]You need to login to perform this action.
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