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.
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When combining cells in parallel, always scrutinize the battery symbols carefully. If one battery is flipped (negative terminal facing A), its emf must be entered as a negative value (e.g., $-E_2$) in the numerator formula.
Concept: • When two non-ideal battery cells are connected completely in parallel, they form a single equivalent cell that can mathematically replace them.
• The total equivalent internal resistance follows the standard parallel resistor formula: $\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$.
• The equivalent electromotive force (emf) is a weighted average determined by the formula: $E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$.
Step 1: Identify parameters from the circuit diagram
Carefully inspecting the provided circuit diagram reveals the connections.
The upper cell has an emf $E_1 = 12\text{ V}$ and an internal resistance $r_1 = 1 \, \Omega$.
The lower cell has an emf $E_2 = 6\text{ V}$ and an internal resistance $r_2 = 0.5 \, \Omega$.
Crucially, the longer vertical lines of both battery symbols (representing the positive terminals) are facing the exact same direction, specifically towards terminal A.
Because they are perfectly aligned with matching polarities, we use standard positive addition in our formula. Step 2: Calculate the equivalent internal resistance
Apply the standard formula for resistors situated in parallel:
\[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \]
Substitute the given resistance values:
\[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{0.5} \]
Recognize that $\frac{1}{0.5}$ is exactly equal to $2$:
\[ \frac{1}{r_{eq}} = 1 + 2 = 3 \]
Invert the result to find $r_{eq}$:
\[ r_{eq} = \frac{1}{3} \, \Omega \approx 0.33 \, \Omega \] Step 3: Calculate the equivalent electromotive force (emf)
Utilize the established equivalent cell formula:
\[ E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_{eq}}} \]
Substitute the individual values rigorously:
\[ E_{eq} = \frac{\frac{12}{1} + \frac{6}{0.5}}{3} \]
Simplify the specific terms within the numerator:
$\frac{12}{1} = 12$
$\frac{6}{0.5} = 12$
Now insert these back into the main calculation:
\[ E_{eq} = \frac{12 + 12}{3} \]
\[ E_{eq} = \frac{24}{3} \]
\[ E_{eq} = 8\text{ V} \]
The equivalent cell operating between points A and B powerfully behaves precisely as an $8\text{ V}$ battery with an internal resistance of $\frac{1}{3} \, \Omega$.
