Circuit Resistance: Find Total Resisitance with Cells Connected in Series

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    Circuit Resistance
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In the discussion about calculating total resistance in a circuit with three cells connected in series, the initial misunderstanding was that the total internal resistance was incorrectly perceived as 0.5 Ω instead of the correct 1.5 Ω. Each cell has an internal resistance of 0.5 Ω, leading to a total internal resistance of 1.5 Ω when three cells are combined. When this is added to the external resistor of 6 Ω, the total resistance of the circuit is correctly calculated as 7.5 Ω. The clarification highlights the importance of accurately interpreting the problem statement. Understanding these calculations is crucial for anyone studying electrical engineering.
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Homework Statement


hi
i am wondering if this is correct. this is the question;
Three cells, each of e.m.f. 1·5 V and internal resistance 0·50 Ω, are connected in series to
make a battery of e.m.f. 4·5V. The battery is connected to a resistor, R, of resistance 6·0 Ω.



Homework Equations


V= i/R


The Attempt at a Solution


TOTAL VOLTAGE = 4.5V
TOTAL RESISTANCE = 6.5 OHMS
but the answer states the total resistnace is 7.5 ohms.
can someone kindly explain how?
thanks
 
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aurao2003 said:
Three cells, each of e.m.f. 1·5 V and internal resistance 0·50 Ω, are connected in series to
make a battery of e.m.f. 4·5V. The battery is connected to a resistor, R, of resistance 6·0 Ω.
Three cells of 0.50Ω + a resistor of 6·0 Ω = ?
 
Fightfish said:
Three cells of 0.50Ω + a resistor of 6·0 Ω = ?

:mad::smile:
silly me.
total internal resistance = 1.5 ohms (3x0.5 ohms)
total resistance = 1.5 +6= 7.5 ohms

thanks a million
 
EACH of the three cells has internal resistance of 0.5Ω

0.5Ω + 0.5Ω + 0.5Ω = 1.5Ω

add that to the resistor of 6Ω connected to the battery and you get 7.5Ω! =)

You probably thought the three resistors had 0.5Ω in total, easy mistake to make in reading the problem.

EDIT: nevermind, seems you got it already.
 
soothsayer said:
EACH of the three cells has internal resistance of 0.5Ω

0.5Ω + 0.5Ω + 0.5Ω = 1.5Ω

add that to the resistor of 6Ω connected to the battery and you get 7.5Ω! =)

You probably thought the three resistors had 0.5Ω in total, easy mistake to make in reading the problem.

EDIT: nevermind, seems you got it already.
yeah. not good for a budding engineer.:wink: thanks
 
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