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Circuit (Network) analysis is the process of finding the voltages across, and the currents through, every component in the network. Consider using in conjunction with tags for the particular analysis technique(s) in question: "nodal-analysis", "mesh-analysis", "superposition", etc.

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Applying superposition in circuit analysis

You have a network that consists of linear elements and independent voltage sources, so why not? You can apply superposition theorem. In your circuit, superposition theorem can be used to calculate cu …
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4 votes
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Biasing circuit with mosfet: am I missing information or misreading it?

The current through both the MOSFETs is the same. Also it is given that both the MOSFETs are in saturation region. Therefore we can write: $$I_{D1} = I_{D2}= I_{D} = k(V_{GS} - V_t)^2 = 20 \mu A $$ on …
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2 votes

How to determine the saturation level when an opamp schmitt trigger starts worknig?

For the sake of theory, we assume an initial state of output. Here in your circuit for inverting schmitt trigger, it is assumed to be \$+V_{sat}.R_2/(R_1+R_2)\$ at the non-inverting terminal, at t= 0, …
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2 votes
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How to calculate voltage of summing op-amp

There are many ways to solve. One way of solving is: Voltage at node b = 0.25 V (Voltage Divider) It should be equal to the voltage at node a too, in ideal op-amp. Hence $$V_a = V_b = 0.25 V$$ Now cu …
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4 votes
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Electrical Engineering - Transformer

Well, \$I >= 0.4 \$ i.e., \$V/Z >= 0.4\$ \$V = 20\$ \$Z =\sqrt{R^2 + X^2}\$, NOT \$(R+X)\$ Solve.
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Voltage divider. Very Basic Question

I presume \$R_A = R_3||R_4\$, and \$R_B = R_2+R_A\$. The voltage across \$R_1\$, is the same as the voltage across the series combination of \$R_2\$ and \$R_A\$. \$R_1\$ doesn't affect the voltage dr …
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2 votes

Diodes circuit analysis problem and voltage calculation

Marking the ground and nodes should help: To calculate \$V_o\$, traverse from ground to \$ A\$, and add the voltages: \$V_o = 2 + V_{AB}\$ where \$V_{AB}\$ is the voltage drop across the 1k\$\Omega\$ …
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1 vote
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Finding Ro while designing an amplifier

Open-circuit voltage is the voltage at output when there is no load connected. So it will be same as the supply voltage at the output circuit. Now, a load \$ R_L\$ is connected and the output resistan …
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1 vote

How do I find the Thevenin resistance and its corresponding voltage?

Is the 16 ohm not part of the resistance calculations anymore You can disregard it, because the other end is open, it doesn't let any current flow through it. do i just add r3 r5 r2 and r4 in serie …
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How this equation of the voltage is attained?

To find voltage between \$A\$ and \$B\$ , traverse from node B to A through any of the paths, and add all the voltage drops along the way. For instance, you can follow this path: P.S: Assuming that t …
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T_hold and T_setup help me understand?

what I don't understand is that why t_skew will be "harmful" when talking about T_hold? Hold violation happens when the data launched by FF1 reaches FF2 "too earlier" than it is supposed to be. Supp …
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13 votes

Why is the efficiency of a half wave rectifier equal to 40.6% and not 50%?

The 'efficiency' they are referring to is Conversion Ratio as I found in the wikipedia article about Rectifiers - Conversion ratio (also called "rectification ratio", and confusingly, "efficiency") η …
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3 votes

Why is there negative sign as well as positive sign with currents flowing into node for KCL

Signs of current in a circuit are just conventions chosen at your convenience. I can say a current of \$1A\$ flows from node A to B. I can also say that current of \$-1A\$ flows from node B to A. Both …
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3 votes

Inconsistency in Values of Laplace Transformed Current when Changing Units

Nothing to do with Laplace. Just some mathematics. I found that you have dimensional inconsistency while you converted units. For instance in this equation :- $$I_1(s)=\frac{20}{s+1}$$ The dimension o …
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1 vote

How to calculate the Impedance of a circuit

First you identify the concerned impedance is across which two points. Let's take the first circuit. The total load impedance seen by the voltage source will be across the voltage source, ie., across …
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