A cell of emf 10 text{ V} and internal resist | vedclass

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(B) The circuit consists of three cells connected in parallel. Let the emfs be $\varepsilon_1 = 10 	ext{ V}$,$\varepsilon_2 = 7 	ext{ V}$,and $\vare

Vedclass · Accepted Answer

$E=2 	ext{ V}, r=0.4 \Omega$

Vedclass · Answer

(B) The circuit consists of three cells connected in parallel. Let the emfs be $\varepsilon_1 = 10 	ext{ V}$,$\varepsilon_2 = 7 	ext{ V}$,and $\varepsilon_3 = 20 	ext{ V}$,with internal resistances $r_1 = 3 \Omega$,$r_2 = 0.6 \Omega$,and $r_3 = 2 \Omega$.Since the third cell is connected with reversed polarity relative to the first two,the equivalent emf $E_{	ext{eq}}$ and equivalent internal resistance $r_{	ext{eq}}$ are given by:$\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} - \frac{\varepsilon_3}{r_3}$$\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{10}{3} + \frac{7}{0.6} - \frac{20}{2} = \frac{10}{3} + \frac{35}{3} - 10 = \frac{45}{3} - 10 = 15 - 10 = 5 	ext{ A}$For parallel combination,the equivalent internal resistance is:$\frac{1}{r_{	ext{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{3} + \frac{1}{0.6} + \frac{1}{2} = \frac{1}{3} + \frac{5}{3} + \frac{1}{2} = 2 + 0.5 = 2.5 	ext{ S}$$r_{	ext{eq}} = \frac{1}{2.5} = 0.4 \Omega$Now,$E_{	ext{eq}} = \left(\frac{E_{	ext{eq}}}{r_{	ext{eq}}}ight) 	imes r_{	ext{eq}} = 5 	imes 0.4 = 2 	ext{ V}$Thus,$E = 2 	ext{ V}$ and $r = 0.4 \Omega$.

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