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You must be trying to find the Thevenin equivalent of the circuit as seen at terminals a,b. If you find the Thevenin equivalents of Vs1, Zs1 and Y2, and the Thevenin equivalents of Vs2, Zs2, Y3, you'll have a fairly simple circuit you can solve knowing Ohm's Law.


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Here's how you might start to simplify it (just a hint): -


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General steps to get started solving many resistor problems, not this problem specifically. Anywhere there are two resistors in direct series, combine them. Redraw the circuit. Anywhere there are two resistors in direct parallel, combine them. Redraw the circuit. Keep doing this until there are no more simple combinations. Now you have a more clear ...


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There are many ways to achieve this in LTspice. What I'll do is step a voltage source from \$0\:\text{V}\$ to \$1\:\text{V}\$, to keep the computation simple. And I'll use the .OP card (not the .TRAN card here.) I can think of other ways, too. Perhaps someone will present a more prosaic approach. But this was easy for me. So here it is: I've arranged for ...


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First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one. In combination with the other answers, my answer is valuable. Well, we are trying to analyze the following circuit: simulate this circuit – ...


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Hints Try redrawing it like this: - 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 ...


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Does it mean that we can't make an equivalent circuit using the Thevenin Theorem? No, you can make an equivalent circuit but, the voltage source will be 0 volts. The equivalent circuit will be just a resistor (450 Ω by my calculation) connected to 0 volts.


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As Hearth points out, an LDR doesn't generate a voltage. It's a highly non-linear light-dependent resistor. I took this "fuzzy" curve from this datasheet, which gives an idea about various LDRs: These curves are not strictly linear in the log-log-domain. But they are close enough that for circuit simulation using a strictly linear log-log-domain ...


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For one thing, your circuit does include a voltage source, you've called it V1 and it's a -5 volt source in series with the LDR. For another thing, the output of a circuit doesn't have to be a voltage. In most cases with LDRs, the output is a resistance, which you can measure by applying a known voltage and measuring the current, or by applying a known ...


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If the sensor is a photoresistor it should be modeled as a resistor, because that's what it is, a resistor that varies depending on how much light falls on it. From the schematic, the LDR forms a voltage divider with Rpol, and it would appear that the voltage across Rpol is what you are reading. The resistance of the LDR changes with light, changing the ...


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