# 3.22 Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value of $\epsilon$

Given

maintained  constant emf of standard cell  = 1.02V, balanced point of this cell = 67.3cm

Now when the standard cell is replaced by another cell with emf = $\varepsilon$. balanced point for this cell = 82.3cm

Now as we know the relation

$\frac{\varepsilon}{l} =\frac{E}{L}$

$\varepsilon =\frac{E}{L}*l=\frac{1.02}{67.3}*82.3=1.247V$

Hence emf of another cell is 1.247V.

## Related Chapters

### Preparation Products

##### Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-