Two cells of emfs 1 V and 2 V and internal resistance and , respectively connected in parallel, gave a current

Question

Two cells of emfs 1 V and 2 V and internal resistance $2 Ω$ and $1 Ω$ , respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\frac{α}{5}$ A. The value of $α$ is _____. [2026]

Smart Achievers · Accepted Answer

(3)

Case-I

$E_{eq} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3} V and r_{eq} = \frac{3}{2} Ω$

$i = \frac{E_{eq}}{r_{eq} + R} = \frac{\frac{5}{3}}{\frac{3}{2} + R} = 1 A$

$\frac{5}{3} = \frac{3}{2} + R \Rightarrow R = \frac{1}{6} Ω$

Case-II

$E_{eq} = \frac{\frac{2}{1} - \frac{1}{2}}{\frac{1}{1} + \frac{1}{2}} = 1 V and r_{eq} = \frac{3}{2}$

$i = \frac{E_{eq}}{r_{eq} + R} = \frac{1}{\frac{3}{2} + \frac{1}{6}} = \frac{3}{5} A \Rightarrow α = 3$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) Case-I Eeq=12+2112+11=5232=53V and req=32Ω i=Eeqreq+R=5332+R=1 A 53=32+R⇒R=16Ω Case-II Eeq=21-1211+12=1 V and req=32 i=Eeqreq+R=132+16=35A⇒α=3

Want Us to Email you About Special Offers & Updates?