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Current Electricity — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsCurrent Electricity3 Marks
Question
In the figure a long uniform potentiometer wireABis having a constant potential gradient along its length. The null points for the two primary cells of emfs $\varepsilon_{1}$ and $\varepsilon_{2}$
connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) $\varepsilon_{1}$/$\varepsilon_{2}$and (ii) position of null point for the cell $\varepsilon_{1}$. How is the sensitivity of a potentiometer increased?

Using Kirchoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 $\Omega$ resistance.Also find the potential difference between A and D.

✓

Answer

  1. $\varepsilon_{1} + \varepsilon_{2}$ = 300k (k is potential gradient in volt/cm)

$\varepsilon_{1} - \varepsilon_{2} = 120 \text{k}$

$\Rightarrow\frac{\varepsilon_{1}}{\varepsilon_{2}} = 7 / 3$

  1. $\varepsilon_{1} + \varepsilon_{2} = 300\text{k}$

$\therefore\varepsilon_{1} + \frac{3}{7}\varepsilon_{1} = 300 \text{k}$

$\Rightarrow\varepsilon_{1} = 210 \text{k}$

Therefore, balancing length for cell $\varepsilon_{1}$ is 210cm.

  1. By decreasing potential gradient.

[Or through Increasing length, reducing potential drop across wire, increasing resistance put in series with the main cell etc.]

Alternate Answer

Applying Kirchhoff's Voltage rule

for loop ABEFA

-9 + 6 + 4 x 0 +2 I = 0

2 I – 3 = 0

$\text{I} =\frac{3}{2}\text{A} = 1.5\text{A}$

For loop BCDEB

3 + I R + 4 x 0– 6 = 0

$\therefore\text{IR} = 3 $

Substituting the value of current I,

$\frac{3}{2}\times\text{R} = 3 $

$\therefore\text{R} = 2\Omega$

Potential difference between A & D

Through path ABCD

+9V – 3V – IR = VAD

$ + 9 - 3 - \frac{3}{2}\times2 = \text{V}_{AD}$

$\Rightarrow\text{V}_{AD} = 3\text{V}$

Alternate Answer

through path AFD

$\frac{3}{2}\times2 =\text{V}_{AD}$

$\Rightarrow\text{V}_{AD} = 3 \text{V}.$

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