Q148 / 160 · AP EAMCET 2024 (21 May Shift 2) · Physics

A cell of emf 1.2 V and internal resistance $2 \Omega$ is connected in parallel to another cell of emf 1.5 V and internal resistance $1 \Omega$. If the like poles of the cells are connected together, the emf of the combination of the two cells is

  • A
    0.8 V
  • B
    3.9 V
  • C
    2.7 V
  • Correct answer
    D
    1.4 V

Solution

$\begin{aligned} & \text { } \mathrm{E}_1=1.2 \mathrm{~V}, \mathrm{r}_1=2 \Omega, \mathrm{E}_2=1.5 \mathrm{~V}, \mathrm{r}_2=1 \Omega \\ & \mathrm{E}_{\text {eq }}=\frac{\frac{\mathrm{E}_1}{\mathrm{r}_1}+\frac{\mathrm{E}_2}{\mathrm{r}_2}}{\frac{1}{\mathrm{r}_1}+\frac{1}{\mathrm{r}_2}}=\frac{\frac{1.2}{2}+\frac{1.5}{1}}{\frac{1}{2}+\frac{1}{1}}=\frac{0.6+1.5}{1.5} \\ & =1.4 \mathrm{~V}\end{aligned}$

Hence, the correct answer is option: 'D'.

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