A potentiometer wire of length 1 m has a resistance of 5 $\Omega$. It is connected to a 8 V battery in series with a resistance of 15 $\Omega$. Determine the emf of the primary cell which gives a balance point at 60 cm.
CBSE DELHI - SET 3 2014
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Current flowing in the potentiometer $\text{I} = \frac{\text{V}}{\text{R} + \text{R}'}$ $=\frac{8.0}{5+15}A=0.4 \ A$ Potential drop across the potentiometer wire V = IR
= 0.4 x 10 V = 2.0 V $K=\frac{V}{l}=\frac{2.0}{1.0}=2.0 \text{ V/m}$Unknown emf $\text{E}=Kl'$
= 2.0 x 0.6 V
= 1.2 V.
= 0.4 x 10 V = 2.0 V $K=\frac{V}{l}=\frac{2.0}{1.0}=2.0 \text{ V/m}$Unknown emf $\text{E}=Kl'$
= 2.0 x 0.6 V
= 1.2 V.
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