To find the value of α \alpha α
Step 1: Initial Parallel Combination of Cells
Given data for the two cells:
Cell 1: EMF E 1 = 1 V E_1 = 1\text{ V} E 1 = 1 V r 1 = 2 Ω r_1 = 2\ \Omega r 1 = 2 Ω
Cell 2: EMF E 2 = 2 V E_2 = 2\text{ V} E 2 = 2 V r 2 = 1 Ω r_2 = 1\ \Omega r 2 = 1 Ω
When two cells with same polarities are connected in parallel, the equivalent EMF (E eq E_{\text{eq}} E eq r eq r_{\text{eq}} r eq
E eq = E 1 r 1 + E 2 r 2 1 r 1 + 1 r 2 E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} E eq = r 1 1 + r 2 1 r 1 E 1 + r 2 E 2
r eq = r 1 r 2 r 1 + r 2 r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} r eq = r 1 + r 2 r 1 r 2
Substituting the given values:
E eq = 1 2 + 2 1 1 2 + 1 1 = 5 2 3 2 = 5 3 V E_{\text{eq}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3}\text{ V} E eq = 2 1 + 1 1 2 1 + 1 2 = 2 3 2 5 = 3 5 V
r eq = 2 × 1 2 + 1 = 2 3 Ω r_{\text{eq}} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\ \Omega r eq = 2 + 1 2 × 1 = 3 2 Ω
Step 2: Determine External Resistance (R R R
The current through the external resistance R R R I = 1 A I = 1\text{ A} I = 1 A I = E eq R + r eq I = \frac{E_{\text{eq}}}{R + r_{\text{eq}}} I = R + r eq E eq
1 = 5 3 R + 2 3 1 = \frac{\frac{5}{3}}{R + \frac{2}{3}} 1 = R + 3 2 3 5
R + 2 3 = 5 3 ⟹ R = 1 Ω R + \frac{2}{3} = \frac{5}{3} \implies R = 1\ \Omega R + 3 2 = 3 5 ⟹ R = 1 Ω
Step 3: Reversing the Polarity of One Cell
If the polarity of one of the cells is reversed, the new equivalent EMF (E eq ′ E_{\text{eq}}' E eq ′
E eq ′ = ∣ E 1 r 1 − E 2 r 2 1 r 1 + 1 r 2 ∣ E_{\text{eq}}' = \left| \frac{\frac{E_1}{r_1} - \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} \right| E eq ′ = r 1 1 + r 2 1 r 1 E 1 − r 2 E 2
E eq ′ = ∣ 1 2 − 2 1 1 2 + 1 1 ∣ = ∣ − 3 2 3 2 ∣ = 1 V E_{\text{eq}}' = \left| \frac{\frac{1}{2} - \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} \right| = \left| \frac{-\frac{3}{2}}{\frac{3}{2}} \right| = 1\text{ V} E eq ′ = 2 1 + 1 1 2 1 − 1 2 = 2 3 − 2 3 = 1 V
The equivalent internal resistance remains unchanged:
r eq = 2 3 Ω r_{\text{eq}} = \frac{2}{3}\ \Omega r eq = 3 2 Ω
Step 4: Calculate the New Current (I ′ I' I ′
The new current passing through the external resistance R R R
I ′ = E eq ′ R + r eq = 1 1 + 2 3 = 1 5 3 = 3 5 A I' = \frac{E_{\text{eq}}'}{R + r_{\text{eq}}} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}\text{ A} I ′ = R + r eq E eq ′ = 1 + 3 2 1 = 3 5 1 = 5 3 A
Comparing I ′ I' I ′ I ′ = α 5 A I' = \frac{\alpha}{5}\text{ A} I ′ = 5 α A
α 5 = 3 5 ⟹ α = 3 \frac{\alpha}{5} = \frac{3}{5} \implies \alpha = 3 5 α = 5 3 ⟹ α = 3