Question 537: Current Electricity

Question

A potentiometer wire of length 1 m is connected in series with $495 \Omega$ resistance and 2 V battery. If $0.2 \mathrm{mV} / \mathrm{cm}$ is the potential gradient, then the resistance of the potentiometer wire is

Accepted Answer

Given:
The total resistance in series is $ R + 495 \, \Omega $.
The battery voltage is 2 V.
The potential gradient along the wire is $ 0.2 \, 	ext{mV/cm} $.
The length of the potentiometer wire, $ L = 1 \, 	ext{m} = 100 \, 	ext{cm} $.
Step 1: Determine the current ($ I $) in the circuit:
$ I = \frac{V}{R_{	ext{total}}} = \frac{2}{R + 495} $
Step 2: Use the potential gradient ($ \phi $) formula:
The potential gradient is given by:
$ \phi = \frac{V}{L} = \frac{IR}{L} = \frac{2R}{(R + 495)} $
Given $\phi = 0.2 \, 	ext{mV/cm} = 0.02 \, 	ext{V/m}$, we can set up the equation:
$ 0.02 = \frac{2R}{(R + 495)} $
Step 3: Solve the equation for $ R $:
$ 0.02 	imes (R + 495) = 2R $
$ 1.98R = 9.9 $
$ R = \frac{9.9}{1.98} = 5 \, \Omega $
Therefore, the resistance of the potentiometer wire is $ 5 \, \Omega $.

Answer

$8 \Omega$

Answer

$6 \Omega$

Answer

$7 \Omega$

Answer

$5 \Omega$

Current Electricity

A potentiometer wire of length 1 m is connected in series with 495 \Omega resistance and 2 V battery. If 0.2 \mathrm{mV} / \mathrm{cm} is the potential gradient, then the resistance of the potentiometer wire is

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