Page 309 - Schaum's Outline of Theory and Problems of Applied Physics

294 DIRECT-CURRENTCIRCUITS [CHAP. 25

The current through the entire circuit is
V 20 V
I = = = 4.17 A
R 4.8
We note that I = I 3 + I 4 and that I 3 = I 1 + I 2 , as they should.

SOLVED PROBLEM 25.12

The Wheatstone bridge (Fig. 25-6) proves an accurate means for determining an unknown resistance R x
with the help of the fixed resistors R 1 and R 2 and the calibrated variable resistor R 3 . Resistance R 3 is
varied until the galvanometer G shows no deflection, a situation described by saying that the bridge is
balanced. Find the value of R x in terms of R 1 , R 2 , and R 3 .

Fig. 25-6

No current passes through the galvanometer when the bridge is balanced, so the same current I 1 passes through
R 1 and R 2 and the same current I 2 passes through R 3 and R x . Also, points A and B must have no potential difference
between them; hence V AC = V BC and V AD = V BD . Since
V AC = I 1 R 1 V AD = I 1 R 2
V BC = I 2 R 3 V BD = I 2 R x
we have the equations

V AC = V BC V AD = V BD
I 1 R 1 = I 2 R 3 I 1 R 2 = I 2 R x
Dividing the last equation in the first column by the last equation in the second column gives

I 1 R 1 I 2 R 3
=
I 1 R 2 I 2 R x
The currents I 1 and I 2 cancel, and so
R 1 R 3 R 2 R 3
= R x =
R 2 R x R 1

EMF AND INTERNALRESISTANCE

The work done per coulomb on the charge passing through a battery, generator, or other source of electric energy
is called the electromotive force,or emf, of the source. The emf is equal to the potential difference across the
terminals of the source when no current flows. When a current I flows, this potential difference is less than the
emf because of the internal resistance of the source. If the internal resistance is r, then a potential drop of Ir
occurs within the source. The terminal voltage V across a source of emf V e whose internal resistance is r when