Question 136: Current Electricity

Question

The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K
1
' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K
1
' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ $\Omega$.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l5wiogku/b967f734-1f81-49aa-9852-c551c2a0a21c/d42f40e0-09c4-11ed-aacd-cd06ee4162a1/file-1l5wiogkv.png?format=png)

Accepted Answer

Shortcut Method :
Internal Resistance of Unknown Battery
$r=\left(\frac{\ell_1-\ell_2}{\ell_2}ight) \mathrm{R}$
Where l
1
means balanced length when key K
1
is open
Where l
2
means balanced length when key K
1
is closed
Here l
1
= 120 cm and l
2
= 80 cm and R = 4 $\Omega $
$ 	herefore $ $r = \left( {{{120 - 80} \over {80}}} ight) 	imes 4$ = 2 $\Omega $
Normal Method :
Original Potentiometer is $	o$
Here, AB is called potentiometer wire. Let length of AB = l and resistance = R
$	herefore$ (1) Current in the circuit (i) = ${{4} \over {R + {R_h}}}$
(2) Potential difference between A and B is
= V
A
$-$ V
B
= $\Delta$V
AB
= i $	imes$ R
(3) ${{\Delta {V_{AB}}} \over {AB}} = {{Potential\,difference} \over {Length}}$ = Potential Gradient
$ \Rightarrow {{\Delta {V_{AB}}} \over l} = {{iR} \over l} = {4 \over {R + {R_h}}} 	imes {R \over l}$ Volt/m
Now, a circuit added to potentiometer wire.
When K
1
is open then circuit look like the following.
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} 	imes 120 = {{4R} \over {(R + {R_h})l}} 	imes 120$ ...... (1)
(2) From external circuit :
V
A
$-$ 1.5 $-$ 0 $	imes$ r  = V
C
$\Rightarrow$ V
A
$-$ V
C
= 1.5V = $\Delta$V
AC
...... (2)
Comparing equation (1) and (2), we get
${{4R} \over {(R + {R_h})l}} 	imes 120 = 1.5$ ..... (3)
When the key K
1
is closed then circuit look like this $	o$
When null point is found at C then no current will flow through the circuit added to the potentiometer wire but i
1
current will flow through circuit created using 1.5V with internal resistance r and  4$\Omega$ resistance.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} 	imes 80 = {{4R} \over {(R + {R_h})l}} 	imes 80$ ....... (4)
(2) From external circuit :
${i_1} = {{1.5} \over {4 + r}}$
${V_A} - 1.5 + {i_1}r = {V_C}$
$ \Rightarrow {V_A} - {V_C} = 1.5 - {i_1}r = 1.5 - {{1.5r} \over {4 + r}}$ ..... (5)
Or we can use this also,
${V_A} - {i_1} 	imes 4 = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {i_1} 	imes 4 = {{1.5} \over {4 + r}} 	imes 4$
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} 	imes 80 = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} 	imes 80 = 1.5 - {{1.5r} \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 	imes {2 \over 3} = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 - 1.5 	imes {2 \over 3}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 	imes {1 \over 3}$
$ \Rightarrow 3r = 4 + r$
$ \Rightarrow 2r = 4$
$ \Rightarrow r = 2\,\Omega $
Other Method :
Now this can be converted to single battery and single resistance using combination of battery rule,
We know,
${E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}$
and ${r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}$
$	herefore$ Here, ${E_{eq}} = {{1.5 	imes 4 + 0 	imes r} \over {4 + r}} = {6 \over {4 + r}}$
and ${r_{eq}} = {{4r} \over {4 + r}}$
$	herefore$ Circuit becomes
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} 	imes 80 = {{4R} \over {(R + {R_h})l}} 	imes 80$ .... (4)
(2) From external circuit :
${V_A} - {6 \over {4 + r}} - 0 	imes {{4r} \over {4 + r}} = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {6 \over {4 + r}}$ ...... (5)
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} 	imes 80 = {6 \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} 	imes 80 = {6 \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 	imes {2 \over 3} = {6 \over {4 + r}}$
$ \Rightarrow 4 + r = 6$
$ \Rightarrow r = 2\,\Omega $
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Current Electricity

Integer / Numeric Answer Type