Practice free →
HomeISC Class 12PhysicsCurrent Electricity › Two cells of emfs $\varepsilon_1$ and $\varepsil…

Two cells of emfs $\varepsilon_1$ and $\varepsilon_2$ with internal resistances $r_1$ and $r_2$ are connected in series with the negative of one joined to the positive of the next. The equivalent emf and internal resistance are

A$\varepsilon_1 + \varepsilon_2$ and $r_1 + r_2$
B$\varepsilon_1 + \varepsilon_2$ and $\dfrac{r_1 r_2}{r_1 + r_2}$
C$\dfrac{\varepsilon_1 + \varepsilon_2}{2}$ and $r_1 + r_2$
D$\varepsilon_1 - \varepsilon_2$ and $r_1 + r_2$
Answer & Solution
Correct answer: A. $\varepsilon_1 + \varepsilon_2$ and $r_1 + r_2$
1. For cells in series aiding, the same current passes through both. 2. The terminal potentials add: $\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$. 3. The internal resistances are in series, so $r_{eq} = r_1 + r_2$. 4. The parallel form (B) and the subtraction (D, which needs opposing cells) do not apply. _Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.15_
Solve this in the app — ISC Class 12 practice & 24k+ MCQs →
Related questions