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Current Through Series of Resistors

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Three resistors having resistances of 1.20ohms , 2.60ohms , and 5.00 ohms are connected in series to a 26.0 V battery that has negligible internal resistance.

1) Find the equivalent resistance of the combination

2) Find the current in through 1.60, 2.50, and 5.00ohms resistors.

3) Find the total current through the battery.

4) Find the power dissipated in 1.60, 2.50, and 5.00ohms resistors

5) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? and why?

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Solution to the current through series of resistors is given.

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1.

The resistors are connected in series.
Therefore the total equivalent resistance is the sum of the individual resistors:

Req = R1+R2+R3 = 1.2 + 2.6 + 5.0 = 8,8 ohms.

2. 3.

In a series connection, conservation of charge dictates that the current (charge per unit time) through all the elements must be the same.

Since all the resistors are connected in series, the current ...

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