A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along with wire?

Given,

Length of potentiometer wire $(l) = 10\;m$

Resistance of potentiometer $(R_p) = 20\; Ω$

Potential of battery $(V_b) = 3V$

Resistance of battery $(R_b) = 10 \; Ω$

Potential Gradient $(\frac{V}{l}) = \;?$

Let $I$ be the current flowing through the circuit.

Then, from Ohm's law,

$V_b = IR$     [where R= Total resistance of the circuit = $R_b + R_p$

or, $3 = I\;*\;(10+20)$

⇒ I = 0.1 A

Since the combination is in series, the current (I) all over the circuit is same.

So, p.d in AB $(V_{AB})$ = I * $R_{b}$ = 0.1 * 20 = 2 V

Now, Potential Gradient $(\frac{V}{l}) = \frac{2}{10}$ = 0.2V/m

Hence, the potential gradient along the wire is 0.2 V/m.

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