(^264) Aptitude Test Problems in Physics
Considering that the resistance RAB of the
circuit satisfies the equation RAB = (11(1 1 + / 2 ),
we obtain
RAB
U (^) U 5 U
11+12 11-F (615) Il = 11 /^2 '
Taking into account the relation obtained above,
we get the following expression for the required
resistance:
15
RAD= 11 R.
3.27. It follows from symmetry considerations
that the initial circuit can be replaced by an equi-
valent one (Fig. 215). We replace the "inner"
triangle consisting of an infinite number of ele-
ments by a resistor of resistance RA B/2, where the
resistance RAB is such that RA B = Rx and RAB =
ap. After simplification, the circuit becomes a sys-
tem of series- and parallel-connected conductors.
In order to find Rx , we write the equation
Rs=
(R
RRx I2 \ R+ RR,/2 1 -1
- RxI2 1 R + 17,121