Question:
Two identical cells having internal resistance $1\Omega$ and emf 6v and 2v are connected in series with external resistance $4\Omega$, find current throught external resistance
Two identical cells having internal resistance $1\Omega$ and emf 6v and 2v are connected in series with external resistance $4\Omega$, find current throught external resistance
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Always check the polarity of series connections. If drawing a circuit diagram isn't provided, standard "in series" usually implies adding voltages.
Updated On: Apr 21, 2026
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Solution and Explanation
Step 1: Determine Equivalent EMF
For series aiding connection: \[ E_{eq} = E_1 + E_2 = 6 + 2 = 8 \text{ V} \]
Step 2: Calculate Internal Resistance
\[ r_{eq} = r_1 + r_2 = 1 + 1 = 2 \Omega \]
Step 3: Total Resistance of Circuit
\[ R_{total} = R_{ext} + r_{eq} = 4 + 2 = 6 \Omega \]
Step 4: Apply Ohm’s Law
\[ I = \frac{E_{eq}}{R_{total}} = \frac{8}{6} \]
Step 5: Simplify
\[ I = \frac{4}{3} \text{ A} \]
Step 6: Final Answer
\[ \boxed{I = \frac{4}{3} \text{ A}} \]
For series aiding connection: \[ E_{eq} = E_1 + E_2 = 6 + 2 = 8 \text{ V} \]
Step 2: Calculate Internal Resistance
\[ r_{eq} = r_1 + r_2 = 1 + 1 = 2 \Omega \]
Step 3: Total Resistance of Circuit
\[ R_{total} = R_{ext} + r_{eq} = 4 + 2 = 6 \Omega \]
Step 4: Apply Ohm’s Law
\[ I = \frac{E_{eq}}{R_{total}} = \frac{8}{6} \]
Step 5: Simplify
\[ I = \frac{4}{3} \text{ A} \]
Step 6: Final Answer
\[ \boxed{I = \frac{4}{3} \text{ A}} \]
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