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  4. Resistance of 100 cm long potentiometer wire is 10 Ω. it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is

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Q. Resistance of $100$ cm long potentiometer wire is $10 \Omega$. it is connected to a battery of 2 volt and resistance R is in series. A source of 10 mV gives null point at 40cm length, then external resistance R is

Current Electricity

Solution:

$V=\left(\frac{E}{R+R_{n}+r}\right) \frac{r}{L} \times \ell$
$\Rightarrow 10 \times 10^{-3}$
$=\frac{2}{(10+R+0)} \times \frac{10}{1} \times 0.4$
$R =790 \Omega$