Charge, current, voltage and power: what the quantities mean

$q$ is the charge, in coulombs, that has crossed a chosen cross-section of the conductor in the direction of the reference arrow; positive charge crossing against the arrow counts as negative. $t$ is time in seconds. $i$ is in amperes. Holds for any conductor and any time variation.

Move a small positive charge $dq$, in C, from a to b. The field does work $dw$ on it, in joules. $\vec{E}$ is the electric field in V/m. $v_{ab}$ is in volts. It is positive when a is at the higher potential, so the + of $v_{ab}$ sits at a. The integral is the same for every path from a to b only where no changing magnetic flux passes between the paths.

The sum runs over every wire that meets the node or crosses the closed surface, with every current referenced into it. It holds when the node is an ideal junction that stores no charge, or the surface encloses only ideal wires and whole elements that are each neutral overall.

Take a closed path that runs along the wires and, at each of the $M$ elements, crosses outside the element from the terminal where you enter it to the terminal where you leave it. $v_k$ is the potential drop from that entry terminal to that exit terminal, in V. The total is zero when no changing flux links this path.

$d$ is the largest dimension of the circuit, in m. $f$ is the highest frequency with significant content, in Hz. $\lambda$ is the distance the signal travels in one period, in m. A real jump, called an edge, takes a finite time: its rise time $t_r$, in s, is the time it takes to go from 10% to 90% of its full change. For a step or a pulse the edge is the fastest change, so compare $t_d$ with $t_r$ instead of with a period: the condition is $t_d\ll t_r$.

A charge $dq$, in C, enters a two-terminal element at its + terminal and leaves at its − terminal. $v$ is the drop from + to −, in V, so by $v=dw/dq$ the charge gives up $dw$, in J, to the element. For $dq=idt$ to be this charge, the arrow of $i$ must point into the + terminal: only then does a positive current carry positive charge from + to −. $p$ is in watts. It holds when the current arrow enters the terminal labelled + for $v$: this is the passive sign convention, used throughout. $p>0$ means the element absorbs energy; $p<0$ means it delivers energy. This is true at each instant and for any two-terminal element.

Call $n_{+}(k)$ the node at the + terminal of branch $k$ and $n_{-}(k)$ the node at its − terminal. With the passive convention, the arrow of $i_k$ leaves node $n_{+}(k)$ and enters node $n_{-}(k)$. A connected piece of the circuit is a set of nodes any two of which are joined by a path of branches. Choose a reference node in each piece. $e_m$, in V, is the potential of node $m$: the sum of drops along any path of branches from node $m$ to its piece's reference, where crossing branch $k$ from $n_{+}(k)$ to $n_{-}(k)$ adds $v_k$ and crossing it the other way adds $-v_k$. One such path followed by another reversed is a closed path, and reversing a path flips the sign of each drop, so KVL makes their sums equal and $e_m$ is well defined. A path from $n_{+}(k)$ that first crosses branch $k$ then has the sum $v_k+e_{n_{-}(k)}$, which must equal $e_{n_{+}(k)}$.

$B$ is the number of branches, $N$ the number of nodes, and $I_m$, in A, the sum of the branch currents leaving node $m$. Collect the terms that multiply each $e_m$. Branch $k$ carries $i_k$ out of node $n_{+}(k)$, giving $+e_{n_{+}(k)}{i}_k$. It carries $i_k$ into node $n_{-}(k)$, which is $-i_k$ leaving it, giving $-e_{n_{-}(k)}{i}_k$. KCL makes the currents leaving each node sum to zero, so every $I_m=0$ and the total power is 0 W.