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The resistance is measured after replacing all voltage- and current-sources with their internal resistances. That means an ideal voltage source is replaced with a short circuit, and an ideal current source is replaced with an open circuit. Resistance can then be calculated across the terminals using the formulae for series and parallel circuits ...
Figure 2: A negative-feedback amplifier. The circuit can be explained by viewing the transistor as being under the control of negative feedback.From this viewpoint, a common-collector stage (Fig. 1) is an amplifier with full series negative feedback.
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This is equivalent to calculating the Thevenin resistance. When there are dependent sources, the more general method must be used. The voltage at the terminals is calculated for an injection of a 1 ampere test current at the terminals. This voltage divided by the 1 A current is the Norton impedance R no (in ohms). This method must be used if ...
When the power source delivers current, the measured voltage output is lower than the no-load voltage; the difference is the voltage drop (the product of current and resistance) caused by the internal resistance. The concept of internal resistance applies to all kinds of electrical sources and is useful for analyzing many types of circuits.
where L, R and C represent inductance, resistance, and capacitance respectively and s is the complex frequency operator = +. This is the conventional way of representing a general impedance but for the purposes of this article it is mathematically more convenient to deal with elastance , D , the inverse of capacitance, C .
Source transformations are easy to compute using Ohm's law.If there is a voltage source in series with an impedance, it is possible to find the value of the equivalent current source in parallel with the impedance by dividing the value of the voltage source by the value of the impedance.
Simplified model for powering a load with resistance R L by a source with voltage V S and resistance R S.. The theorem was originally misunderstood (notably by Joule [4]) to imply that a system consisting of an electric motor driven by a battery could not be more than 50% efficient, since the power dissipated as heat in the battery would always be equal to the power delivered to the motor when ...