Current Electricity

JEE Main level, numerical answer, medium

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Question 168 of 380

Two cells of emfs 11 V and 22 V and internal resistance 2 Ω2\,\Omega and 1 Ω1\,\Omega, respectively connected in parallel, gave a current of 11 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be α5\dfrac{\alpha}{5} A. The value of α\alpha is __________.

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Answer: 3

For two cells connected in parallel with the same polarity, the equivalent emf EeqE_{eq} and equivalent internal resistance reqr_{eq} are given by:

Eeq=E1r2+E2r1r1+r2E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}

req=r1r2r1+r2r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}

Substituting the given values (E1=1E_1 = 1 V, r1=2 Ωr_1 = 2\,\Omega, E2=2E_2 = 2 V, r2=1 Ωr_2 = 1\,\Omega):

Eeq=1×1+2×22+1=53E_{eq} = \dfrac{1 \times 1 + 2 \times 2}{2 + 1} = \dfrac{5}{3} V

req=2×12+1=23 Ωr_{eq} = \dfrac{2 \times 1}{2 + 1} = \dfrac{2}{3}\,\Omega

The current II through the external resistance RR is 11 A:

I=EeqR+reqI = \dfrac{E_{eq}}{R + r_{eq}}

1=53R+231 = \dfrac{\dfrac{5}{3}}{R + \dfrac{2}{3}}

R+23=53⇒R=1 ΩR + \dfrac{2}{3} = \dfrac{5}{3} \Rightarrow R = 1\,\Omega

When the polarity of one cell is reversed, the new equivalent emf Eeq′E'_{eq} becomes:

Eeq′=∣E1r2−E2r1∣r1+r2=∣1×1−2×2∣3=33=1E'_{eq} = \dfrac{|E_1 r_2 - E_2 r_1|}{r_1 + r_2} = \dfrac{|1 \times 1 - 2 \times 2|}{3} = \dfrac{3}{3} = 1 V

The equivalent internal resistance remains the same, req=23 Ωr_{eq} = \dfrac{2}{3}\,\Omega.

The new current I′I' through the external resistance RR is:

I′=Eeq′R+req=11+23=153=35I' = \dfrac{E'_{eq}}{R + r_{eq}} = \dfrac{1}{1 + \dfrac{2}{3}} = \dfrac{1}{\dfrac{5}{3}} = \dfrac{3}{5} A

Given that the new current is α5\dfrac{\alpha}{5} A, we have:

α5=35⇒α=3\dfrac{\alpha}{5} = \dfrac{3}{5} \Rightarrow \alpha = 3

Answer: 33

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