Question

Two cells of emfs 12 V and 6 V are connected in parallel as shown in the figure. Their internal resistances are \(1\,\Omega\) and \(0.5\,\Omega\) respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.

Answer 1

Given

\[ E_1=12\,V,\qquad r_1=1\,\Omega \]

\[ E_2=6\,V,\qquad r_2=0.5\,\Omega \]

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Step 1: Calculate the Equivalent emf

For two cells connected in parallel (with like terminals connected), the equivalent emf is

\[ E=\frac{\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}} {\dfrac{1}{r_1}+\dfrac{1}{r_2}} \]

Substituting the given values,

\[ E= \frac{\dfrac{12}{1}+\dfrac{6}{0.5}} {\dfrac{1}{1}+\dfrac{1}{0.5}} = \frac{12+12}{1+2} = \frac{24}{3} =8\,V \]

Hence,

\[ \boxed{E_{\rm eq}=8\,V} \]

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Step 2: Calculate the Equivalent Internal Resistance

The equivalent internal resistance of cells in parallel is

\[ r=\frac{r_1r_2}{r_1+r_2} \]

Substituting the values,

\[ r=\frac{1\times0.5}{1+0.5} =\frac{0.5}{1.5} =\frac{1}{3}\,\Omega \]

Therefore,

\[ \boxed{r_{\rm eq}=\frac{1}{3}\,\Omega\approx0.33\,\Omega} \]

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Final Answer

Equivalent emf:

\[ \boxed{8\,V} \]

Equivalent internal resistance:

\[ \boxed{\frac{1}{3}\,\Omega\approx0.33\,\Omega} \]