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schematic and original question

I tried to take the 5 Ω resistor as Rl but still couldn't find the answers as given in the green highlight. I'm pretty new at this and can only manage solving straightforward circuits.

I managed to reduce the circuit and obtain the Rth value, but don't know which node to take in order to find Vth. I am also confused about which portion of the circuit doesn't have a voltage drop, I presume it's the 100 V and 10 Ω resistor.

enter image description here

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    \$\begingroup\$ Please edit your question to show the equations you've written. This will allow us to show you where you mistake is, and otherwise your question is considered off-topic for lack of (apparent) effort. \$\endgroup\$
    – Null ♦
    Jan 4, 2022 at 17:47
  • \$\begingroup\$ Edit: I managed to reduce the circuit - did you reduce the circuit following the hints in my answer? \$\endgroup\$
    – Andy aka
    Jan 4, 2022 at 18:20
  • \$\begingroup\$ no i just saw it now, i have a question though, i understand the resistors but could u explain why does the 50 v turn into 20v ? \$\endgroup\$
    – swaloo
    Jan 4, 2022 at 18:21
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    \$\begingroup\$ That's fundamental when converting a voltage source that has both series resistance and parallel resistance. It forms a potential divider and the open circuit Thevenin voltage is 20 volts. Your picture is wrong in this one respect. However, it doesn't affect your evaluation of the Thevenin resistance because voltage sources are set to zero. \$\endgroup\$
    – Andy aka
    Jan 4, 2022 at 18:23
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    \$\begingroup\$ Show us your attempt to find the Thevenin voltage. Explain what you mean by "which node to take". Explain why you think a "portion of the circuit doesn't have voltage drop". \$\endgroup\$
    – Elliot Alderson
    Jan 4, 2022 at 18:31

1 Answer 1

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Hints

Try redrawing it like this: -

enter image description here

This would be the first stage reduction of the problem by redrawing and shifting things around. The next stage would involve combing the remaining 50 volt source with the new 20 volt source - again this is a significant reduction in order to make the problem easier to solve.

However, if you can't follow what I've done above in the pictures, you may need to speak to your professor. Good luck; it's fairly easy from here.

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