Q.49. Two cells of emfs 1V and 2V and internal resistance and , respectively connected in parallel,

Q: What is the correct answer for this exam Physics Q49?

The correct answer is D. Case 1: Cells in parallel, same polarity. Cell 1: $E_1 = 1V$, $r_1 = 2\Omega$. Cell 2: $E_2 = 2V$, $r_2 = 1\Omega$. External resistance $R$. Current $I = 1A$. For parallel combination of cells, $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$ and $r_{eq} = \frac{1}{1/r_1 + 1/r_2}$. $E_{eq} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{0.5 + 2}{0.5 + 1} = \frac{2.5}{1.5} = \frac{5/2}{3/2} = \frac{5}{3} V

Q: Q.49. Two cells of emfs 1V and 2V and internal resistance and , respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then…

The correct answer is D. Case 1: Cells in parallel, same polarity. Cell 1: $E_1 = 1V$, $r_1 = 2\Omega$. Cell 2: $E_2 = 2V$, $r_2 = 1\Omega$. External resistance $R$. Current $I = 1A$. For parallel combination of cells, $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$ and $r_{eq} = \frac{1}{1/r_1 + 1/r_2}$. $E_{eq} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{0.5 + 2}{0.5 + 1} = \frac{2.5}{1.5} = \frac{5/2}{3/2} = \frac{5}{3} V

Question

Q.49. Two cells of emfs 1V and 2V and internal resistance  and , respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be  A. The value of  is _______.

FirstInTest · Accepted Answer

Case 1: Cells in parallel, same polarity. Cell 1: $E_1 = 1V$, $r_1 = 2\Omega$. Cell 2: $E_2 = 2V$, $r_2 = 1\Omega$. External resistance $R$. Current $I = 1A$. For parallel combination of cells, $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$ and $r_{eq} = \frac{1}{1/r_1 + 1/r_2}$. $E_{eq} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{0.5 + 2}{0.5 + 1} = \frac{2.5}{1.5} = \frac{5/2}{3/2} = \frac{5}{3} V$. $r_{eq} = \frac{1}{1/2 + 1/1} = \frac{1}{1.5} = \frac{1}{3/2} = \frac{2}{3} \Omega$. The total current is $I = \frac{E_{eq}}{r_{eq} + R}$. $1 = \frac{5/3}{2/3 + R}$. $1 = \frac{5}{2 + 3R}$. $2 + 3R = 5$. $3R = 3$. $R = 1\Omega$. Case 2: Polarity of one cell is reversed. Let's reverse $E_1$. So $E_1 = -1V$. $E_{eq}' = \frac{-1/2 + 2/1}{1/2 + 1/1} = \frac{-0.5 + 2}{1.5} = \frac{1.5}{1.5} = 1 V$. $r_{eq}' = r_{eq} = \frac{2}{3} \Omega$. The current through the external resistance $R=1\Omega$ is $I' = \frac{E_{eq}'}{r_{eq}' + R} = \frac{1}{2/3 + 1} = \frac{1}{5/3} = \frac{3}{5} A$. The question states the current will be $\frac{\alpha}{5} A$. So $\frac{\alpha}{5} = \frac{3}{5}$. This implies $\alpha = 3$.

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1

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