Thus the energy supplied by the source is qV, while that dissipated by the resistors is Electrical potential energy can be described by the equation PE = qV, where q is the electric charge and V is the voltage. This equation is based on the conservation of energy and conservation of charge. The sum of these voltages equals the voltage output of the source that is, So the voltage drop across R 1 is V 1 = IR 1, that across R 2 is V 2 = IR 2, and that across R 3 is V 3 = IR 3. Another way to think of this is that V is the voltage necessary to make a current I flow through a resistance R. According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).
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