P ) in a circuit is the rate at which ENERGY is absorbed or produced within a circuit.2) A source of energy such as a VOLTAGE will produce or deliver power while the connected load absorbs it.NOTE: The higher their value or rating in watts the more electrical power they are likely to consume.Mathematically power is represented as:P = V × IP = V2/RP = I2 × RPower triangle is shown belowTellegen's theoremPower delivered by source = Power absorbed by all elementsCalculation:Given amplifier supply 12 VAC voltage of 10 V peak then SIGNAL is V(t) = Vmsin(ωt) or V(t) = Vmcos(ωt)Vm = 10 VLoad resistance is 1 KΩRMS value of the sinusoidal signal is\({V_{rms}} = \frac{{{V_m}}}{{\sqrt 2 }}\)Power delivered by the amplifier isP = 12 V × (10 + 10) mAP = 240 mWPower absorbed by the load\(P = \frac{{V_{rms}^2}}{R}\)\({P_L} = \frac{{{{\left( {\frac{10}{{\sqrt 2 }}} \right)}^2}}}{{1\;K{\rm{\OMEGA }}}}\)PL = 50 mWDissipated power isPdis = 240 mW – 50 mWPdis = 190 mW

"> P ) in a circuit is the rate at which ENERGY is absorbed or produced within a circuit.2) A source of energy such as a VOLTAGE will produce or deliver power while the connected load absorbs it.NOTE: The higher their value or rating in watts the more electrical power they are likely to consume.Mathematically power is represented as:P = V × IP = V2/RP = I2 × RPower triangle is shown belowTellegen's theoremPower delivered by source = Power absorbed by all elementsCalculation:Given amplifier supply 12 VAC voltage of 10 V peak then SIGNAL is V(t) = Vmsin(ωt) or V(t) = Vmcos(ωt)Vm = 10 VLoad resistance is 1 KΩRMS value of the sinusoidal signal is\({V_{rms}} = \frac{{{V_m}}}{{\sqrt 2 }}\)Power delivered by the amplifier isP = 12 V × (10 + 10) mAP = 240 mWPower absorbed by the load\(P = \frac{{V_{rms}^2}}{R}\)\({P_L} = \frac{{{{\left( {\frac{10}{{\sqrt 2 }}} \right)}^2}}}{{1\;K{\rm{\OMEGA }}}}\)PL = 50 mWDissipated power isPdis = 240 mW – 50 mWPdis = 190 mW

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A sinusoidal voltage of 1 V peak is applied at the input of an amplifier which operates at ± 12 V power supplies. A sinusoidal voltage of 10 V peak is available across a 1 kΩ load at the output of the amplifier. The input current to the amplifier from the sinusoidal source is negligible. If the amplifier draws a current of 10 mA from each of the two power supplies, the power dissipated in the amplifier is

Electronics & Communication Engineering Op-Amp And Its Applications in Electronics & Communication Engineering . 7 months ago

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Concept:1) Electrical Power, ( P ) in a circuit is the rate at which ENERGY is absorbed or produced within a circuit.2) A source of energy such as a VOLTAGE will produce or deliver power while the connected load absorbs it.NOTE: The higher their value or rating in watts the more electrical power they are likely to consume.Mathematically power is represented as:P = V × IP = V2/RP = I2 × RPower triangle is shown belowTellegen's theoremPower delivered by source = Power absorbed by all elementsCalculation:Given amplifier supply 12 VAC voltage of 10 V peak then SIGNAL is V(t) = Vmsin(ωt) or V(t) = Vmcos(ωt)Vm = 10 VLoad resistance is 1 KΩRMS value of the sinusoidal signal is\({V_{rms}} = \frac{{{V_m}}}{{\sqrt 2 }}\)Power delivered by the amplifier isP = 12 V × (10 + 10) mAP = 240 mWPower absorbed by the load\(P = \frac{{V_{rms}^2}}{R}\)\({P_L} = \frac{{{{\left( {\frac{10}{{\sqrt 2 }}} \right)}^2}}}{{1\;K{\rm{\OMEGA }}}}\)PL = 50 mWDissipated power isPdis = 240 mW – 50 mWPdis = 190 mW

Posted on 30 Oct 2024, this text provides information on Electronics & Communication Engineering related to Op-Amp And Its Applications in Electronics & Communication Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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