Question

Two identical cells of emf $1.5V$ each connected in parallel provide a supply to an external circuit consisting of two resistors of $17\Omega$ each joined in parallel. A very high resistance volt meter reads the terminal voltage of the cells to be $1.4V$. What is the internal resistance of each cell?

• A. $1.21\Omega$
• B. $0.3\Omega$
• C. $0.4\Omega$
• D. $0.5\Omega$

Answer 1

The correct option is A $1.21\Omega$

The resultant resistance of $R_1$ and $R_2$ is given by,

$R=\frac{R_1\times R_2}{R_1+R_2}=\frac{17\times 17}{17+17}=8.5$

Using Kirchhoff's lar for loop $ABCDEA$

$I{R}_{eq}+I_1{R}_{1}0V_{emf}=0\to \left(1\right)$

Using Kirchhoff's lar for loop $ABCDFA$

$I{R}_{eq}+R_2{I}_2-V_{emf}=0\to \left(2\right)$

From $\left(2\right)$ and $\left(1\right)$

$I_1=\frac{v_I{R}_{eq}{r}_1}{,}{I}_2=\frac{V_{emp}-I{R}_{eq}}{r_2}$

and $I=I_1+I_2=\frac{v_{emf}-I{R}_{eq}}{r_1}+\frac{V_{emf}-I{R}_{eq}}{r_2}$

$\therefore r_1=r_2$

$⟹I=\frac{2(V_{emf}-I{R}_{eq})}{r}$

$Ir=2(V_{emf}-I{R}_{eq})$

$⟹Ir+2I{R}_{eq}=2V_{emf}$

$I=\frac{2V_{emf}}{r+2R_{eq}}$

Also, $V=I{R}_{eq}$

$⟹I=\frac{V}{R_{eq}}=\frac{2V_{emf}}{r+2R_{eq}}$

Substituting value of $R_{eq},V$ and $V_{emf}$ we get

$\frac{1.4}{8.5}=\frac{1.5}{(8.5+r/2)}$

$⟹r=1.21\Omega$