Question C asks how's to calculate the current over the led and 3k resistor? Can someone please help me understand how to do it? 
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1\$\begingroup\$ I(3k)=I(470Ω)+I(LED) and Vled=2V so V(3k)=18V got it? \$\endgroup\$– D.A.S.Commented Dec 18, 2018 at 22:39
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1\$\begingroup\$ In the future please use a photo editor like GIMP or Photoshop to remove the stuff we do not need to see. You can use the cropping tool to select just the section we need to see. \$\endgroup\$– user105652Commented Dec 18, 2018 at 23:01
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\$\begingroup\$ Convert the original voltage source and the resistor divider into a new Thevenin voltage source and Thevenin output resistance. Subtract the diode voltage from the Thevenin voltage source value, then divide this resulting voltage by the Thevenin resistance you computed to get the LED current. \$\endgroup\$– jonkCommented Dec 18, 2018 at 23:43
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1 Answer
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20V supply and 2V across the LED leaves how much voltage across the 3K Ω resistor?
With that voltage, use ohms law to calculate the current through the 3K Ω resistor.
Calculate the current for a 470 Ω resistor at 2v.
There is more current flowing through the 3K Ω than the 470 Ω resistor.
Where would that excess I3K - I470Ω current go?
answered Dec 18, 2018 at 23:27
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\$\begingroup\$ I'll upvote since the OP did not. \$\endgroup\$– user103380Commented Dec 19, 2018 at 16:28
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\$\begingroup\$ Ye I've basically done that myself. I just wasn't sure what I was doing was right. Appreciate the help! Also give me chance to reply KingDuken. \$\endgroup\$ Commented Dec 19, 2018 at 19:21