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I thought the Zener diode (break down voltage = 6V) will not be active here. That is it will not start acting as voltage regulator here since potential difference across 4 kΩ is 4 V if not in connected with diode in parallel. So after connecting in parallel it should still not be acting as 6 V, it will remain 4 V and current flowing through circuit will be 10/1000 = 0.01 A.

enter image description here0.01A?

(Answer given was using active state (10 - 6) V / 6 kΩ = 0.667 mA.)

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  • \$\begingroup\$ Welcome! I tried to clear up your question but please edit ”10-6/6K\ohm” and clarify it. \$\endgroup\$
    – winny
    Apr 16 at 11:15
  • \$\begingroup\$ Edited thanks. .. \$\endgroup\$
    – Orion_Pax
    Apr 16 at 11:35
  • \$\begingroup\$ For all future reference, it’s kohm or kΩ, not “K\ohm”. \$\endgroup\$
    – winny
    Apr 16 at 11:54
  • \$\begingroup\$ Sure okay winny \$\endgroup\$
    – Orion_Pax
    Apr 16 at 14:31

1 Answer 1

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You are correct except for

... and current flowing through circuit will be 10/1000 = 0.01 A.

The resistance is 10 kΩ so \$ I = \frac {10}{10k} = 1 \ \text{mA}\$.

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