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I am currently simulating an AC mains fuse blown indicator circuit using Proteus 8. However, I can't seem to find out the correct calculation as to why the Voltage and Current across the 550 Ohm resistor and the fuse is the value shown in the image below.The voltage that appeared on the multimeter I believe is the RMS value?

Can anyone help me to find the value attained above, please?

Thank you in advance

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  • \$\begingroup\$ What is L2? Can you put a voltmeter across L2 and update the image? As this is an AC circuit, is it possible for you to show the phases of different voltages(especially across L2) ? \$\endgroup\$ – Ashutosh Dec 20 '17 at 9:34
  • \$\begingroup\$ The current in the circuit, and thus the voltages across different components depend on the type of load L2. \$\endgroup\$ – Ashutosh Dec 20 '17 at 9:36
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Not without knowing more about L2. But from the values I calculate it to be about 26 Ohm. Then the I and V make sense as V R7 = 550/(550+26)*220 = ~210V. And I is 220/(550+26)= 0.38A. The resistance of the rest of the circuit (D4,D5 etc) is high compared to the resistance of the bulb and disappears in the accuracy of the measured current value.

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  • \$\begingroup\$ I see that explains a lot. The L2 is indeed set 24 ohms in the simulation which I forgot to take note of. May I also ask, why is the current flowing to Emitter side of the PNP transistor zero? Is it because there is current flowing into the Base? Since for the PNP transistor to conduct current from Emitter to Collecter, current must be leaving the Base instead \$\endgroup\$ – Michael Andrian Dec 20 '17 at 10:45
  • \$\begingroup\$ Your LED is in the emitter and needs 2.5V thus your Vb must be around 3.1V below the voltage at the top before the LED lights up. Which will happen if the fuse blows but then Vbe will be around 220V and probably your transistor will warn you using smoke signals. \$\endgroup\$ – Oldfart Dec 20 '17 at 10:58

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