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

Fig. 220

Solutions 273

5-6 as shown in Fig. 219. Then it follows from sym-
metry considerations that there is no current
through this wire. Therefore, the central junction
can be removed from the initial circuit, and we
arrive at the circuit shown in Fig. 220. By hy-
pothesis,

= R13 = R 34 =R24= r,

r

1/-2
Let U be the voltage between points 1 and 2.
Then the amount of heat liberated in the conductor
1-2 per unit time is

U2
Q12= •

From Ohm's law, we obtain the current through

the conductor 3-4:

U

I8 4 -

r (V2+3) •

The amount of heat liberated in the conductor 8-4

per unit time is

Q34 = /14r

r (1/2+3) 2 *

R15= R25= R36 = R46=

U 2

18-0771