Circuit Analysis with Two Batteries and an Ammeter

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The discussion focuses on calculating the current reading on an ammeter when a switch is closed in a circuit with two batteries and resistors. The calculated electromotive force (emf) of the left battery is approximately 27 V, leading to an initial current calculation of 0.5 A using the formula I = V/R. There is confusion regarding the influence of the second battery and other resistors on the ammeter's reading, suggesting that the current calculation may overlook these components. The principle of superposition is recommended for analyzing the circuit, emphasizing that each battery can be considered independently if they have no internal resistance. Understanding the voltage across the 50-ohm resistor and the properties of ideal voltage sources is crucial for accurate analysis.
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Homework Statement


http://imgur.com/6Cds4YF

What will the ammeter read when the switch is closed?

Homework Equations


V = IR

The Attempt at a Solution


I already calculated the emf of the battery one the left to be about 27 V.
It seems that the answer to the question is as simple as using V = IR where V is from the battery.
So I = V/R = 25/50 = 0.5 A.

I don't understand why the other battery to the left and the other resistors don't affect the current through the ammeter though. Wouldn't using the numbers for calculating the current, assume that the only item in the circuit is a 25 V battery and one 50 ohm resistor?
 
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What is the voltage across the 50 ohm resistor? What factors affect what that voltage is?
 
Use principle of superposition, one battery at a time.
 
If the battery has no internal resistance it behaves like an ideal voltage source. Think about the properties of an ideal voltage source.
 
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