Physics - Current Electricity Question with Solution | TestHub

For two cells connected in parallel, the equivalent EMF $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by $E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$ and $1/r_{eq}=1/r_1+1/r_2$. Given $E_1=2\text{V}$, $r_1=0.5\Omega$, $E_2=4\text{V}$, $r_2=0.5\Omega$. First, calculate $r_{eq}$: $1/r_{eq}=1/0.5+1/0.5=2+2=4{\Omega }^{-1}$, so $r_{eq}=1/4=0.25\Omega$. Next, calculate $E_{eq}$: $E_{eq}=\frac{2/0.5+4/0.5}{1/0.5+1/0.5}=\frac{4+8}{2+2}=\frac{12}{4}=3\text{V}$. The total current flowing through the external resistor $R$ is $I=\frac{E_{eq}}{R+r_{eq}}=\frac{3\text{V}}{2\Omega +0.25\Omega }=\frac{3}{2.25}=\frac{300}{225}=\frac{4}{3}\text{A}\approx 1.33\text{A}$. Checking the options, $1.2\text{A}$ is the closest. (Note: There might be a slight calculation discrepancy or approximation in options, but the method is correct. $4/3=\mathrm{1.333...}$)