Question

What is the potential drop between points $A$ and $C$ in the following circuit?
Resistances $1\Omega$ and $2\Omega$ are represent the internal resistance of the respective cells.

• A. 1.75 V
• B. 2.25 V
• C. $\frac{5}{4}V$
• D. $\frac{4}{5}V$

Answer 1

Answer: (B)

Emf's $E_1$ and $E_2$ are opposing each other. Since,
$E_2>E_1$ ,so current will flow from right to left

Current in the circuit
$i=\frac{Netemf}{TotalResistance}=\frac{E_2-E_1}{R+r_1+r_2}$
Given, $R=5\Omega$ , $r_1=1\Omega$ , $r_2=2\Omega$
$E_1=2V$ and $E_2=4V$
$\therefore i=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25A$
The potential drop between points A and C,
$V_A-V_C=E_1+i{r}_1$
$=2+0.25\times 1$
$=2.25V$