Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 265

Solving this equation, we obtain

R AB=Rx

R (117 — 1) ap —1)

-" (^3) — •
3
3.28. It follows from symmetry consideraions
that if we remove the first element from the circuit,
the resistance of the remaining circuit between
points C and D will be RcE, = kRAB. Therefore,
the equivalent circuit of the infinite chain will
A

R

I

R2 [II I " {Ad III

A B

Fig. 216

have the form shown in Fig. 216. Applying to this
circuit the formulas for the resistance of series-
and parallel-connected resistors, we obtain

R 1 + R 2 kRAB

RI —R2+ V-11 + °RIR

BAR

2

3.29. The potentiometer with the load is equivalent

to a resistor of resistance

R (^) RR 12 5
RI — (^) — R.
2 R -1-- RI2 6
Hence the total current in the circuit will be
U 6 U
/ 2 —
(5/6) R — 5 R •
R AB — R2+ kRAB
Solving the quadratic equation for BA B, we obtain
(in particular, for k = 1/2)