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Question 41 of 42

Q.Two cells of emfs ε1\varepsilon_1 & ε2\varepsilon_2 and internal resistances r1r_1 & r2r_2 respectively are connected in parallel. Obtain expressions for the equivalent.

(i) resistance and
(ii) emf of the combination
Yanam BieapCBSE Class XII Board 2018Subjective· 3mImportance★★★★★
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1req=1r1+1r2\dfrac{1}{r_{eq}}=\dfrac{1}{r_1}+\dfrac{1}{r_2} and εeqreq=ε1r1+ε2r2\dfrac{\varepsilon_{eq}}{r_{eq}}=\dfrac{\varepsilon_1}{r_1}+\dfrac{\varepsilon_2}{r_2}.

Concept. For cells in parallel the same terminal potential difference VV appears across both. Let the total current drawn be I=I1+I2I=I_1+I_2.

Why these formulas. For each cell V=ε−IrV=\varepsilon-I r, so I1=ε1−Vr1I_1=\dfrac{\varepsilon_1-V}{r_1} and I2=ε2−Vr2I_2=\dfrac{\varepsilon_2-V}{r_2}.

Step-by-step.

I=I1+I2=ε1−Vr1+ε2−Vr2=(ε1r1+ε2r2)−V(1r1+1r2).I=I_1+I_2=\frac{\varepsilon_1-V}{r_1}+\frac{\varepsilon_2-V}{r_2}=\left(\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\right)-V\left(\frac{1}{r_1}+\frac{1}{r_2}\right). …

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