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Look at the circuit. • It has a source voltage: v(t) = 149 Cos (2 t 640 t) · And it has a load such as a motor or solenoid - which is modelled by a resistor and pure inductance in series. Look carefully at the sinusoidal signals as they appear on the right-hand side of the scope. They are the same frequency, but they are not in phase (the peaks do not line up in time). The upper signal is the voltage across the load, the lower signal is the current through the load. The phase shift between the two is caused by the electrical property of inductance according to its terminal behaviour. Notice that the load voltage on the top peaks before the load current, we say the voltage across the load leads the current through it. In a purely resistive circuit, they would be in phase. The time difference between the two signals is 375 us. You need to work out the phase difference in degrees and radians. To do this you need to know: • how a signals frequency and period interrelate • how many degrees and radians there are in 1 period of a sinusoid • how degrees and radians interrelate The phase difference in degrees is deg The phase difference in radians rad The current signal would be written as i(t) I Cos (2 n 640 t Ф) RUN / Stop Electric Potential OPower OCharge Source Simulation Speed R1 680mQ Inductive V1 640HZ Load Charge Flow SL1 2.7mH Power Brightness Current Circuit: 149V 13.7A

Look at the circuit. • It has a source voltage: v(t) = 149 Cos (2 t 640 t) · And it has a load such as a motor or solenoid - which is modelled by a resistor and pure inductance in series. Look carefully at the sinusoidal signals as they appear on the right-hand side of the scope. They are the same frequency, but they are not in phase (the peaks do not line up in time). The upper signal is the voltage across the load, the lower signal is the current through the load. The phase shift between the two is caused by the electrical property of inductance according to its terminal behaviour. Notice that the load voltage on the top peaks before the load current, we say the voltage across the load leads the current through it. In a purely resistive circuit, they would be in phase. The time difference between the two signals is 375 us. You need to work out the phase difference in degrees and radians. To do this you need to know: • how a signals frequency and period interrelate • how many degrees and radians there are in 1 period of a sinusoid • how degrees and radians interrelate The phase difference in degrees is deg The phase difference in radians rad The current signal would be written as i(t) I Cos (2 n 640 t Ф) RUN / Stop Electric Potential OPower OCharge Source Simulation Speed R1 680mQ Inductive V1 640HZ Load Charge Flow SL1 2.7mH Power Brightness Current Circuit: 149V 13.7A

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Look at the circuit. • It has a source voltage: v(t) = 149 Cos (2 n 640 t) · And it has a load such as a motor or solenoid - which is modelled by a resistor and pure inductance in series. Look carefully at the sinusoidal signals as they appear on the right-hand side of the scope. They are the same frequency, but they are not in phase (the peaks do not line up in time). The upper signal is the voltage across the load, the lower signal is the current through the load. The phase shift between the two is caused by the electrical property of inductance according to its terminal behaviour. Notice that the load voltage on the top peaks before the load current, we say the voltage across the load leads the current through it. In a purely resistive circuit, they would be in phase. The time difference between the two signals is 375 us. You need to work out the phase difference in degrees and radians. To do this you need to know: • how a signals frequency and period interrelate · how many degrees and radians there are in 1 period of a sinusoid • how degrees and radians interrelate The phase difference in degrees is deg The phase difference in radians rad The current signal would be written as i(t) I Cos (2 n 640 t %D Ф) RUN / Stop Electric Potential OPower OCharge Source Simulation Speed R1 680m2 Inductive V1 640HZ Load Charge Flow SL1 2.7mH Power Brightness Current Circuit: 149V 13.7A
Transcribed Image Text:Look at the circuit. • It has a source voltage: v(t) = 149 Cos (2 n 640 t) · And it has a load such as a motor or solenoid - which is modelled by a resistor and pure inductance in series. Look carefully at the sinusoidal signals as they appear on the right-hand side of the scope. They are the same frequency, but they are not in phase (the peaks do not line up in time). The upper signal is the voltage across the load, the lower signal is the current through the load. The phase shift between the two is caused by the electrical property of inductance according to its terminal behaviour. Notice that the load voltage on the top peaks before the load current, we say the voltage across the load leads the current through it. In a purely resistive circuit, they would be in phase. The time difference between the two signals is 375 us. You need to work out the phase difference in degrees and radians. To do this you need to know: • how a signals frequency and period interrelate · how many degrees and radians there are in 1 period of a sinusoid • how degrees and radians interrelate The phase difference in degrees is deg The phase difference in radians rad The current signal would be written as i(t) I Cos (2 n 640 t %D Ф) RUN / Stop Electric Potential OPower OCharge Source Simulation Speed R1 680m2 Inductive V1 640HZ Load Charge Flow SL1 2.7mH Power Brightness Current Circuit: 149V 13.7A
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