TABLE 19-1.2 Inductors Value (H) and Resistance Vr=1 V VZ Vi=5 V VL f=100 Hz Vac=1 V VAC = 5 V-p f = 15 kHz Vac=1 Vp-p VAC = 5 Vp-p pop 47MH 15mW 100 224
3. Connect the circuit of Fig. 19-1.5, but do not in- sert the inductor. R = 100 2 A VDC = 1 V ve 11.30 10 4 Bo Fig. 19-1.5 Inductor in a series circuit (with current-limiting resistor). 4. Apply 1 V DC across points A-B without the in- ductor in the circuit. Turn off the power supply and insert the inductor. Turn on the power supply and measure the voltage across the inductor. Record the value in Table 19-1.2. 5. Increase the applied voltage to 5 V. Measure and record the value in Table 19-1.2. 6. Replace the inductor with the next value. Mea- sure the voltage across the inductor at 1 V and at 5 V, and record the results in Table 19-1.2. 7. Repeat step 6 for the remaining inductors. 8. Connect the circuit of Fig. 19-1.6, but do not connect the inductor. Adjust the power supply so that 1 Vp-p at 100 Hz appears across A-B. PP AQ
Fig. 19-1.6 Inductor in a series with AC source. р 9. Measure the voltage across the inductor at 1 Vp-p and 100 Hz. Record the value in Table 19-1.2. 10. Increase the voltage to 5 Vp-p. Measure and record the value in Table 19-1.2. 11. Increase the frequency to 15 kHz. Measure the inductor voltage at 1 Vp-p and at 5 Vp-p applied. Record the results in Table 19-1.2. 12. Replace the inductor with the next value of in- ductor, and repeat the measurements in steps 9, 10, and 11. Then repeat this procedure for each value of inductor you have.
TABLE 19-1.2 Inductors Value (H) and Resistance Vr=1 V VZ Vi=5 V VL f=100 Hz Vac=1 V VAC = 5 V-p f = 15 kHz Vac=1 Vp-p V
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TABLE 19-1.2 Inductors Value (H) and Resistance Vr=1 V VZ Vi=5 V VL f=100 Hz Vac=1 V VAC = 5 V-p f = 15 kHz Vac=1 Vp-p V
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