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 _____________ $$\Omega$$.

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

Integer / Numeric Answer Type