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2

cfr provided the basic answer that will get you up and running quickly with your existing code. But I provide this answer so you can see some ideas you might find useful in the future. Here's another way of drawing the circuit without manually specifying the coordinates. It's a little bit more typing at the beginning, but if you decide later to change the ...


2

Based on Harish Kumar's comment: \documentclass[tikz,border=5pt]{standalone} \usepackage{circuitikz} \begin{document} \begin{circuitikz} [american voltages, baseline=(current bounding box.center)] \ctikzset { label/align = straight } \draw (0,0) to[V=$V_{Th}$] (0,2) to[R=$R_{Th}$] (2.5,2) to[short,i=$I$, -o] (4,2) to[short] ...


0

Because the circuits have different heights, the simplest way seems to be manual correction of their positions, for example: \documentclass[11pt]{article} \usepackage{circuitikz} \usepackage{tikz} % for flowcharts' \begin{document} %\begin{center} \begin{tabular}{ccc} \begin{tabular}{c} \begin{circuitikz}[american ...


5

You can do very easily to such things with the circuits libraries of TikZ. A similar approach to add additional pseudo-anchors to a rectangular node can be found in Tikz surrounding box with automatically drawn border “ports”. A more proper way would be to define special anchors with an extra shape so that you would have anchors like .west 1, .west 2, etc. ...


3

Some foreach, two fit nodes and some labels. A nicer example in TeXample.net. \documentclass[tikz, border=2mm]{standalone} \usetikzlibrary{fit} \begin{document} \begin{tikzpicture}[c/.style={circle,fill, minimum size=4pt, inner sep=0pt, outer sep=0pt}] \foreach \i [count=\xe from 0, count=\xo from 4, evaluate={\ni=int(2*\i)}, ...


5

Here's how I would do it. Instead of manually adjusting the position of the labels, use the anchors of the transistors and position the label nodes left or right of these. to paths leading nowhere (++(0,0)) can be used to add filled or open marks at the transistor anchors. I also avoided manually positioning the transistor "C" by using the calc library to ...


2

You can change it with voltage/distance from node= key. I have also added voltage/bump b=20pt,voltage/european label distance=20pt, just in case you want to change them too. \documentclass{article} \usepackage[american,siunitx]{circuitikz} \usetikzlibrary{positioning} \usepackage{siunitx} \ctikzset{voltage/bump b=20pt,voltage/european label ...


3

A possible solution is following. I've copied pmos definition from pgfcirctripoles.sty file to preamble and commented out last lines which draw the circle. That's all. \documentclass{article} \usepackage{tikz} \usepackage{circuitikz} \makeatletter \pgfcircdeclaremos{pmos}{ \anchor{S}{ \northeast } \anchor{source}{ ...


2

You can use fit library as suggested by cfr. Here is an example: \documentclass[11pt]{article} \usepackage{circuitikz} \usetikzlibrary{fit,shapes.geometric} %% shapes.geometric needed for ellipse \begin{document} \begin{center} \begin{circuitikz}[american voltages, american currents, european resistors] \draw (0,0) to[V, l=$50$V] (0,2) to[short, -*] ...


2

Use to[I, i_>=$4$A] (7.5,2) (7.5, 0): \documentclass[11pt]{article} \usepackage{circuitikz} \usepackage{tikz} \begin{document} \begin{center} \begin{circuitikz}[american voltages, american currents, european resistors] \draw (0,0) to[V, l=$50V$] (0,2) to[short, -*] (1.5,2) to[R, l=$5\Omega$] (3.5,2) to[R, ...


2

Use ,baseline=(current bounding box.center) in the options of circuitikz \documentclass[11pt]{article} \usepackage[margin=1in]{geometry} \usepackage{graphicx} \usepackage[english]{babel} \usepackage{circuitikz} \usepackage{tikz} \usetikzlibrary{shapes.geometric, arrows} \begin{document} \begin{center} \begin{circuitikz}[american ...


1

Straight labeling can be achieved with the \ctikzset macro: \documentclass[a4paper]{article} \usepackage{circuitikz} \begin{document} \begin{center} \begin{circuitikz}[american voltages] \ctikzset { label/align = straight } \draw (0,0) to[R, l=$R_c$, o-o] (4,0) % The voltage source to[R, l=$R_a$, o-o] (2,-3) to[R, l=$R_b$, o-o] (0,0) ...



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