Question 440: 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 'K1' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K1' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ $$\Omega$$.

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]
$$ \Rightarrow 1.5 	imes {2 \over 3} = {6 \over {4 + r}}$$
$$ \Rightarrow 4 + r = 6$$
$$ \Rightarrow r = 2\,\Omega $$

Current Electricity

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