Calculating Current in a Series Circuit with Multiple Resistors and a Battery

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The discussion centers on calculating current in a series circuit consisting of three resistors (2, 3, and 5 ohms) connected to a 6V battery. The total resistance is determined to be 10 ohms, leading to a current calculation of 6V / 10 ohms, which equals 0.6 A. It is confirmed that the current through each resistor and the battery remains constant in a series circuit. The concept of an ideal battery, which has no internal resistance, is also clarified.

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Three resitors values 2, 3 and 5 ohms are connected in series whith a battery of 6 V. Caluclate the current through each resistor and the current through the battery.

My answers, (which are wrong) lol.

Current through battery = V =IXR = 6/0 = 0 A

This is my other problem, is the current through a series circuite the same if so do i add all the resistor values up then work out the current? This is so simple to what I am used to its confusing me! Help any one... :cry: :bugeye:
 
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Resistors in series add resistance values like this:

R_\textrm{total} = R_1 + R_2 + ... + R_N

In this case, the total resistance is 10 ohms. The current through each resistor must be the same, since they are in series. A 6V battery driving a 10 ohm load will deliver 6V / 10 ohms of current. The current through each resistor is thus just 6V / 10 ohms.

The current through the battery is, in fact, the hard part of this question. For the sake of most easy electronics problems, you can assume the battery is an "ideal" voltage source with unspecified internal mechanism. Current can be assumed to flow through it in the same way that it flows through the resistors. The ideal battery contributes no resistance to the circuit, but somehow generates a potential difference and causes the electrons to move through it.

- Warren
 
Thank you very very much warren, great help... :-)
 

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