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akch2002, Engineer
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(d) The circuit shown in FIGURE 3 is part of the interface

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(d) The circuit shown in FIGURE 3 is part of the interface of a relay
output module. Ib is 1 mA and VCC is 9 V. The relay requires a
minimum of 50 mA to energise.
Complete the values of the assumptions listed below in order to
• voltage across R1 – 5V- 0.6V = 4.4V
• value of R1 – 4.4/1Ma = 4.4 kΩ
• voltage across the relay coil
• voltage across R2
• value of R2
• collector of current Ic.
Logic '1' = 5 V
Logic '0' = 0V
Transistor forward current gain hfe = 50 (i.e. 50 × 1mA = 50Ma)
LED current = 10 mA
LED voltage drop at 10 mA = 2.4V
Base/emitter voltage = 0.6V
Collector emitter voltage when transistor is on = 1 V
Thanks for your request. Kindly upload the figure. Further, a price of $7.69 for such questions is low. I would request for a bonus of $10.
Customer: replied 4 years ago.
I am working on this question and will try to reply asap. Do you agree to pay bonus as requested?
Customer: replied 4 years ago.
yes no problem
I will reply shortly.

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