Question 383: Current Electricity

Question

Two cells of emfs 1 V and 2 V and internal resistances $2 \Omega$ and $1 \Omega$, respectively, are connected in series with an external resistance of $6 \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $\mathrm{I}_2$. The value of $\left(\frac{\mathrm{I}_1}{\mathrm{I}_2}\right)$ is $\frac{x}{3}$. The value of $x$ is___________.

Accepted Answer

Series Connection:
Total emf = 1 V + 2 V = 3 V.
Total resistance = 2 ÃÂ© + 1 ÃÂ© + 6 ÃÂ© = 9 ÃÂ©.
Current $$I_1 = \frac{	ext{emf}}{	ext{resistance}} = \frac{3}{9} = 	frac{1}{3}	ext{ A}.$$
Parallel Connection:
Equivalent emf:

$$\varepsilon_{	ext{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{	frac{1}{2} + 	frac{2}{1}}{	frac{1}{2} + 1} = \frac{5}{3}	ext{ V}.$$
Equivalent internal resistance of the cells:

$$r_{	ext{int,eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3}\,\Omega.$$
Total resistance = external 6 ÃÂ© + internal $	frac{2}{3}$ ÃÂ© = $	frac{20}{3}$ ÃÂ©.
Current $$I_2 = \frac{\varepsilon_{	ext{eq}}}{	ext{total resistance}} = \frac{5/3}{20/3} = 	frac{1}{4}	ext{ A}.$$
Finally,

$$\frac{I_1}{I_2} = \frac{	frac{1}{3}}{	frac{1}{4}} = \frac{4}{3},$$

so $x = 4\,$.

Current Electricity

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