Question

An external resister is connected across a series combination of two identical cells, each of emf E and internal resistance r. If the current in the circuit is I , the terminal potential difference across the cell is

Answer 1

The terminal voltage difference =V1 -V2

= V=IR

Answer 2

The terminal potential voltage across cell 1 is $r_1$  [ $\dfrac{V}{r_1+r_2+R}$ ]   volt

The terminal potential voltage across cell 2 is  $r_2$  [ $\dfrac{V}{r_1+r_2+R}$ ]   volt  

Explanation:

Given as :

For a series combination of two identical cells, each of emf E and internal resistance r .

The current flowing through the circuit = I amp

Let The terminal potential difference across cells = V

According to question

A series combination of two identical cells, each of emf E and internal resistance r . The supply current is I amp and voltage is V

From figure

For The series combination of cells

The equivalent Emf = Emf across cell 1 +Emf across cell 2 + ........... + Emf across cell n

i.e  $E_e_q$  = $E_1$  +  $E_2$ + ..............+ $E_n$

And , Equivalent Resistance = $R_e_q$

i.e  $R_e_q$  = $R_1$  +  $R_2$ + ..............+ $R_n$

So, The current in the circuit will flow equally through each resistor

Current = I = $\dfrac{V}{R_e_q}$

For circuit shown in fig  with two cell and external resistor R

i.e   $R_e_q$  = $r_1$  +  $r_2$   + R

∴           I = $\dfrac{V}{r_1+r_2+R}$

As series resistance have different voltage

So,   EMF across cell 1 = $E_1$  = $r_1$ I

i.e                                     $E_1$  = $r_1$  [ $\dfrac{V}{r_1+r_2+R}$ ]

Similarly

EMF across cell 2 = $E_2$  = $r_2$ I

i.e                                     $E_2$  = $r_2$  [ $\dfrac{V}{r_1+r_2+R}$ ]

Hence, The terminal potential voltage across cell 1 is $r_1$  [ $\dfrac{V}{r_1+r_2+R}$ ]   volt

And  The terminal potential voltage across cell 2 is  $r_2$  [ $\dfrac{V}{r_1+r_2+R}$ ]   volt  

Answer