Solving for Terminal Voltage and Current in a Battery Circuit

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A battery with a terminal voltage of 12.0 V and an internal resistance of 2.0 Ω is analyzed when a 1.7 Ω resistor is connected. The total circuit resistance is calculated as 3.7 Ω, leading to a current of approximately 3.24 A through the circuit. The voltage drop across the internal resistance is determined using Ohm's law, which affects the terminal voltage. The resulting terminal voltage is less than the battery's emf due to this drop. The discussion emphasizes treating internal resistance as a series component to find the correct values.
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A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 1.7 Ω resistor is connected across the battery terminals, what is the terminal voltage and what is the current through the 1.7 Ω resistor?


Can someone point me in the right direction?
I know I have to use V = E - Ir because there is a internal resistance.
But I am not sure how to start...

Thanks in advance.
 
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Treat the internal resistance as a series resistor, find the total circuit current.

What is the voltage drop across the internal resistance? What do you think that will do to the terminal voltage?
 
The terminal voltage is less than the emf because there is a voltage drop across the internal resistance.

I am still confused.
 
Last edited:
As said earlier treat internal resistance as a series resistor

net resistance= R+r = 2+1.7

current= E/R = 12/3.7

terminal voltage = E - Ir
 
thank you so much!
 
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