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Well, $\text{R}_3$, $\text{R}_4$, $\text{R}_5$ and $\text{R}_6$ are in parallel and they are in series with $\text{R}_1$ and $\text{R}_2$. simulate this circuit – Schematic created using CircuitLab So the total resistance is: \text{R}_\text{total}=\text{R}_1+\text{R}_2+\frac{1}{\frac{1}{\text{R}_3}+\frac{1}{\text{R}_4}+\frac{1}{\text{R}... 5 Just for a definitional answer: Two resistors are in parallel if each terminal of resistor A are connected to the terminals of resistor B Two resistors are in series if only one terminal of resistor A is connected to a terminal of resistor B Just because it's worth saying: it doesn't matter how you draw them, it only matters how you connect them. As for ... 4 Current and voltage are linear, but power is quadratic. So you can't apply superposition on power directly. But you can find either voltage or current using superposition and then work power: Let $V_1$ be the voltage across $R$ when current source is open. Let $V_2$ be the voltage across $R$ when voltage source is shorted. Then by superposition, ... 2 Resistors R3, R4, R5 and R6 are connected in parallel. And this circuit is connected in series to R1 and R2. 2 There are plenty of ways to determine these values. I personally like superposition a lot because it is already part of the divide-and-conquer strategy promoted by the fast analytical techniques or FACTs. Here, superposition applies because we obviously deal with a linear circuit. We thus determine the voltage $V_{th2}$ across the 4-$k\Omega$ resistor ... 2 As @beccaboo mentions, this is possible with a single op-amp, and in this particular case, the numbers fall out nicely, making it relatively simple. Where this circuit fails in practice will be its input impedance being set by the opamp itself rather than the resistor network. Of course, the same concept would apply if we were to add a resistor divider on ... 2 In a switching converter, the input voltage and the output current are considered perturbations. The temperature or the air pressure could also be considered perturbations depending on the operating environment. The system strives to reject these disturbances for the purpose of keeping the output variable, $V_{out}$ for instance, in regulation. In order to ... 2 The speed of the electrons have nothing to do with the "height" of the voltage. Static electricity is a prime example. You can have a 4MV charge on insulated object and none of the electrons are moving. 1 You have been here long enough, I think, to be aware of the schematic editor. It's helpful (saves me time, anyway) if you bother, because the parts get labeled and it's easier to ask you clarifying questions. (But if all you have is a cell phone, it may be the case that you can only add pictures.) That said, here's the schematic I wish you'd drawn: simulate ... 1 I know that the answer is 24 volts by traversing from B to D to C to A adding up voltages of resistors, but should I not be considering the contribution of other elements in the circuit, for example, the 10 V source? Yes it seems confusing but it isn't. If there's something uniquely in series with a current source then that "something" has no effect; ... 1 It depends on your circuit, but there could be a lot of differences. First: With diodes you will not be able to pull the output below the diodes forward voltage, which could be 0.7V. With a resistor divider, you could go lower in voltage. Second: The diodes can handle a push-pull configuration for X1/X2 because if one signal line is driving VCC, the ... 1 The diode AND provides excellent pulldown, and provides slow pullup but a strong "High" level that is limited only by leakage currents in downstream diodes. 1 The current through the base-emitter junction of a transistor is governed by the diode equation: $i_e = I_S \left(e^\frac{v_{be}}{n\ V_T} - 1\right)$. At room temperature, $V_T \simeq 26\mathrm{mV}$. So unless your peak-peak input voltage is significantly smaller than $26\mathrm{mV}$, you'll get distortion. On the bright side, you have lots of ... 1 What is the need for R3? It may be required to pull the output up if/when the inputs are 'floating' (open-circuit). To understand its function we would need to know more about the circuit. How does the following circuit ANDs the 2 signals X1 and X2? The output corresponding to case 2 and 3 in truth table fall under indeterminate LOGIC level of IC ... 1 View this as a voltage divider. If either "in" is low, a robust High" will not occur. Only if both inputs are "high" will a robust "high" be output. This gate depends upon accurate thresholding. These circuits cannot be cascaded. 1 Your problem is for your Opamp. As I see, your RC value is 100nF and 1MOhms is 0.1 seconds, and your plot is a nice 1/0.1s+1 plot. So everything is OK based on your circuit. if you want to increase speed, lower C or R to get a better speed. Your 500 ms delay is a 5 time of time constant of this circuit, which is also shows me that it's a first order ... 1 The math comes from G/(1 + G * H) which is the easily revived transfer function [ math] for a certain type of feedback loop [ formal system concept == math]. 1 When you assume the diode states than you need to ensure that there are no contradictions in the resulting circuit configuration. By just looking at the I-V characteristics of diode it seems like the ON state is possible but the resulting circuit does not follow KVL. For the current direction you assumed, the voltage drop across the loop is \V_1 - V_c > ... 1 Yes, if you find that the short circuit current is negative then the short circuit current actually flows "upward" (assuming the standard way problems are drawn with two relevant nodes aligned one above the other). And if the short circuit current actually flows upward then the Norton source must be oriented pointing down, so that the current will again flow ... 1 Actually R3,R4,R5,R6 are in parallel combination as the voltage across the resistors is same and the combination and R1,R2 are in series as the current will be same through them 1 Your differential equations are correct, but it's probably easier to use the Laplace transform. Determining the impedances of both main branches:\small Z_1=\frac{R_1+R_2+sC_1R_1R_2}{1+sC_1R_1} \small Z_2=\frac{R_3+R_4+sC_2R_3R_4}{1+sC_2R_3} $$Then the currents through each of these branches are:$$\small I_1(s)=\frac{Z_2}{Z_1+Z_2}I(s) \...