Question

# A potentiometer wire has length 4 m and resistance $$8\Omega$$. The resistance that must be connected in series with the wire and an accumulator of e.m.f 2V, so as to get a potential gradient 1 mV per cm on the wire is

A
44Ω
B
48Ω
C
32Ω
D
40Ω

Solution

## The correct option is C $$32\Omega$$the total resistance after adding resistance R in series with potentiometer wire =$$8\Omega+R\Omega$$the current in the circuit will be $$i=V/R_{eq}=\frac{2V}{R+8}$$the potential drop across potentiometer wire will be $$iR_{potentiometer}=\frac{2V}{R+8}\times 8$$the potential gradient along length 4 m is given as $$=V_potetiometer=\frac{\frac{2V}{R+8}\times 8}{4}$$ the gradient is given as $$1mV/cm=.1 V/m$$$$\frac{\frac{2V}{R+8}\times 8}{4}=.1$$$$R=32\Omega$$Physics

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