# Series Circuit Problem Solving Let the voltage output of the battery and resistances in the parallel connection in [link] be the same as the previously considered series connection: , , , and . (d) Calculate the power dissipated by each resistor.

Let the voltage output of the battery and resistances in the parallel connection in [link] be the same as the previously considered series connection: , , , and . (d) Calculate the power dissipated by each resistor.

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So the voltage drop across is , that across is , and that across is .This gives Discussion for (b) Current for each device is much larger than for the same devices connected in series (see the previous example).A circuit with parallel connections has a smaller total resistance than the resistors connected in series.Strategy and Solution for (c) The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage.Thus, This is consistent with conservation of charge.Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded).For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. (See [link](b).) To find an expression for the equivalent parallel resistance , let us consider the currents that flow and how they are related to resistance.Choosing , and entering the total current, yields More complex connections of resistors are sometimes just combinations of series and parallel.These are commonly encountered, especially when wire resistance is considered.Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are , , and .Conservation of charge implies that the total current produced by the source is the sum of these currents: The terms inside the parentheses in the last two equations must be equal.

## Comments Series Circuit Problem Solving

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