circuit {ResistorArray}

Mensurates a circuit given potentials of some nodes and current flow into the others

Description

Given a conductance matrix, a vector of potentials at each node, and a vector of current inputs at each node (NA being interpreted as “unknown”), this function determines the potentials at each node, and the currents along each edge, of the whole circuit.

Usage

circuit(L, v, currents=0, use.inverse=FALSE, give.internal=FALSE)

Arguments

L

Conductance matrix

v

Vector of potentials; one element per node. Elements with NA are interpreted as “free” nodes, that is, nodes that are not kept at a fixed potential. The potential of these nodes is well defined by the other nodes in the problem. Note that such nodes must have current inputs (which may be zero) specified by argument currents

currents

Vector of currents fed into each node. The only elements of this vector that are used are those that correspond to a node with free potential (use NA for nodes that are at a specified potential). The idea is that each node has either a specified voltage, or a specified current is fed into it; not both, and not neither.

Observe that feeding zero current into a node at free potential is perfectly acceptable (and the usual case)

use.inverse

Boolean, with default FALSE meaning to use solve(A,b) and TRUE meaning to use solve(A), thus incurring the penalty of evaluating a matrix inverse, which is typically to be avoided if possible.

The default option should be faster most of the time, but YMMV

give.internal

Boolean, with TRUE meaning to return also a matrix showing the node-to-node currents, and default FALSE meaning to omit this

Value

Depending on the value of Boolean argument give.internal, return a list of either 2 or 4 elements:

potentials

A vector of potentials. Note that the potentials of the nodes whose potential was specified by input argument v retain their original potentials; symbolically all(potentials[!is.na(v)] == v[!is.na(v)])

currents

Vector of currents required to maintain the system with the potentials specified by input argument v

internal.currents

Matrix showing current flow from node to node. Element [i,j] shows current flow from node i to node j. This and the next two elements only supplied if argument give.internal is TRUE

power

The power dissipated at each edge

total.power

Total power dissipated over the resistor network

Note

The SI unit of potential is the “Volt”; the SI unit of current is the “Ampere”

Author(s)

Robin K. S. Hankin

See Also

resistance

Examples

#reproduce first example on ?cube:
v <- c(0,rep(NA,5),1,NA)
circuit(cube(),v)
circuit(cube(),v+1000)

#  problem: The nodes  of a skeleton cube are at potentials
#  1,2,3,... volts.  What current is needed to maintain this?  Ans:
circuit(cube(),1:8)


#sanity check: maintain one node at 101 volts:
circuit(cube(),c(rep(NA,7),101))

#now, nodes 1-4 have potential 1,2,3,4 volts.  Nodes 5-8 each have one
#Amp shoved in them.  What is the potential of nodes 5-8, and what
#current is needed to maintain nodes 1-4 at their potential?
# Answer:
v <- c(1:4,rep(NA,4))
currents <- c(rep(NA,4),rep(1,4))
circuit(cube(),v,currents)

# Now back to the resistance of a skeleton cube across its sqrt(3)
# diagonal.  To do this, we hold node 1 at 0 Volts, node 7 at 1 Volt,
# and leave the rest floating (see argument v below); we
# seek the current at nodes 1 and 7
# and insist that the current flux into the other nodes is zero
# (see argument currents below):

circuit(L=cube(),v=c(0,NA,NA,NA,NA,NA,1,NA),currents=c(NA,0,0,0,0,0,NA,0))

# Thus the current is 1.2 ohms and the resistance (from V=IR)
# is just 1/1.2 = 5/6 ohms, as required.