# Double-Subscript Notation for Voltages

I have read from some book that: "The voltage quantity symbol V sometimes has subscripts to designate the two points to which the voltage corresponds. If the letter a designates one point and b the other, and if W joules of work are required to move Q coulombs from point b to a then, $V_{ab}=\frac{W}{Q}$. Note that the first subscript is the point to which the charge is moved"

While Wikipedia (http://en.wikipedia.org/wiki/Double_subscript_notation) says: "$V_{CB}$ means the "$V$" from $C$ to $B$."

Can you help me with the true meaning of the first subscript?

• There is no 'true meaning' as the notation was used by several disciplines before attempts to standardize it were made. The first letter is generally a starting reference and the last the end reference, but rules always seem to have exceptions such as Vdd or Vcc for power supplies. – JIm Dearden Jun 21 '13 at 8:08
• – RedGrittyBrick Jun 21 '13 at 9:35

I've taught it this way before to give a mental picture of double subscript notation.

$V_{ab}$ is the voltage measured by a voltmeter when the red (positive) lead is connected to node a and the black (negative) lead is connected to node b.

A single subscript, $V_a$, is measured with red to node a and black to node 0 ("ground").

It then follows that: $V_{ab} = V_a - V_b$

• That's very logical – Script Kitty Sep 2 '18 at 16:08

Your Wikipedia link refers to engineering notation. Electrostatic theory is not engineering and the notations are not necessarily always the same (see Jim Dearden's comment).

Having said that, in this case the two notations are actually the same. In electrostatics, the voltage at a point 'a' is defined as the work done in moving a unit charge from infinity to point 'a', ie up the potential gradient. So for two points 'a' and 'b', $V_{ab}$ is still the voltage at 'a' with respect to 'b'.

So in both cases $V_{ab}$ is positive if 'a' is at a higher (more positive) voltage than 'b' and negative if 'a' is at a lower voltage than 'b'. In the case of a bipolar transistor, an NPN device will normally have it's collector at a higher voltage than it's emitter so $V_{CE}$ will be positive, whilst for a PNP transistor the reverse is true so $V_{CE}$ will be negative.

• @ MikeJ-UK Thank you very much! May I ask, can I view $V_{ab}$ as Q coulombs moved from b to a or from a to b? – user7777777 Jun 21 '13 at 17:44
• @user7777777 - From 'b' to 'a' as your book says. – MikeJ-UK Jun 24 '13 at 9:03