Introduction to circuit analysis Of the many electronic components, perhaps the
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Introduction to circuit analysis
Of the many electronic components, perhaps the most common is the resistor. Energy is released, generally in the form of heat, when an electric current flows through a resistor. From the perspective of a circuit, resistors consume energy and expel it to the atmosphere. The energy consumed by a resistor typically comes from a power supply, which is also called a voltage source. Voltage sources add energy to the circuit, and in fact the definition of the volt unit is imparting one joule of energy for every coulomb of charge passing through a particular potential difference, or voltage, i.e. the unit V = J/C. Be careful to distinguish between a potential difference (synonymous with the term voltage) V and the unit volt, which his abbreviated V. Note that variable names are italicized and unit symbols are not. For example a particular voltage source might supply five volts of potential, or V = 5 V. A current is the flow of charge, which is generally measured in amperes where A = C/s indicating that an ampere is the rate at which charge flows.
Energy, or electric potential, is lost when current flows through a resistor. This relationship is governed by Ohm’s law where V=iR
with i representing current, R resistance, and V the voltage drop through the resistor. Resistance is typically measured in ohms, which is abbreviated ?. In order to be consistent with Ohm’s law, the unit ohm must satisfy
?=V/A=J*s/C^2
Unfortunately, the unit of an ohm does not have an intuitively understandable meaning (voltage is energy per charge, and current is charge per time), and most problems can be solved understanding than V = A??.
Some circuits involving multiple resistors can be analyzed simply by combining resistors in series and parallel. Two resistors in series, like those in Figure 1, can be replaced with a single resistor whose resistance is the sum of the original two resistors. Combining resistors, as in Figure 2, is more complicated where the parallel resistances R1 and R2 can be replaced with a resistance
R = R1R2/R1+R2
The circuit in Figure 3 contains a 5 V power supply and various resistors. In order to calculate the current required from the power supply, combine the resistors in series and parallel to reduce the circuit to one containing the voltage source and a single resistor. Then apply Ohms law to compute the required current. Be sure to use consistent units in all calculations.
R1 R2 R+R2 Figure 1: Equivalent circuits when combining resistors in series. R,R R1 R +R2 Figure 2: Equivalent cireuits when combining resistors in parallel.Explanation / Answer
500 ohm and 2kohm are in parallel connection, their equivalent is R' then
1/R' = 1/500 + 1/2000
R' = 400 ohm
now 1k ohm and 400 ohm are in series.
R" = 400 + 1000 = 1400 ohm
now R" and 4kohm are in parallel,
1/R'" = 1/1400 + 1/4000
R"' = 1037.04 ohm
now R'" and 3kOhm are in series,
Req = R"' + 3000 = 4037.04 ohm
current through battery I = 5/ (4037.04) = 1.24 x 10^-3 A = 1.24 mA
current through 3 kohm = 1.24 mA
voltage = I R = 1.24mA x 3kohm = 3.72 Volt
voltage across 4 kohm and R" = 5 - 3.72 = 1.28 V
current through 4 k ohm = 1.28 / 4 = 0.32 mA
current through 1 k ohm = 1.24 - 0.32 = 0.92 mA
current through 500 ohm = 0.92 x 2000 / (2000 + 500) = 0.74 mA
voltage across 500 ohm = 500 x 0.74 =0.37 V
current through 2kohm = 0.92 - 0.74 = 0.18 mA
voltage across 2 k = 0.18 x 2 = 0.36 V
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