. A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 15 V battery and an external resistance 40 Ω. A secondary cell of emf E in the secondary circuit is balanced by 240 cm long potentiometer wire. The emf E of the cell is
Added by Arun B.
Step 1
The potential gradient is given by the formula: Potential gradient = Voltage / Length In the primary circuit, the voltage is 15 V and the length is 10 m. Plugging these values into the formula, we get: Potential gradient = 15 V / 10 m = 1.5 V/m Show more…
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A potentiometer wire of length $10 \mathrm{~m}$ and resistance $20 \Omega$ is connected in series with a $15 \mathrm{~V}$ battery and an external resistance $40 \Omega$. A secondary cell of emf $E$ in the secondary circuit is balanced by $240 \mathrm{~cm}$ long potentiometer wire. The $\operatorname{emf} E$ of the cell is (a) $2.4 \mathrm{~V}$ (b) $1.2 \mathrm{~V}$ (c) $2.0 \mathrm{~V}$ (d) $3 \mathrm{~V}$
A potentiometer wire of length 200cm has a resistance of 20?. It is connected in series with a resistance of 10? and an accumulator of emf 6V having negligible internal resistance. A source of 2.4V is balanced against a length L of the potentiometer wire. The value of L is?
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