Two cells with internal resistances $r_1$ and $r_2$ are connected across the load resistance R as shown in the figure. The values of $r_1,r_2$ and R are $(1/2)Ω, (1/3)Ω$ and $(9/5)Ω$, respectively. The current through R is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7/10 A
Given: $E_1=2\ \text{V},\ E_2=1\ \text{V},\ r_1=\frac{1}{2}\ \Omega,\ r_2=\frac{1}{3}\ \Omega,\ R=\frac{9}{5}\ \Omega$
Equivalent emf of two sources in parallel:
$E_{\text{eq}}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}} =\frac{\frac{2}{\frac{1}{2}}+\frac{1}{\frac{1}{3}}}{\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{3}}} =\frac{4+3}{2+3} =\frac{7}{5}\ \text{V}$
Equivalent internal resistance:
$r_{\text{eq}}=\frac{r_1 r_2}{r_1+r_2} =\frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}+\frac{1}{3}} =\frac{\frac{1}{6}}{\frac{5}{6}} =\frac{1}{5}\ \Omega$
Total resistance:
$R_{\text{tot}}=R+r_{\text{eq}}=\frac{9}{5}+\frac{1}{5}=\frac{10}{5}=2\ \Omega$
Current through $R$:
$I=\frac{E_{\text{eq}}}{R_{\text{tot}}} =\frac{\frac{7}{5}}{2} =\frac{7}{10}\ \text{A}$