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

Solutions (^287)
By hypothesis, the current required for melting
the first wire must exceed 10 A. Therefore,
k • 441 (Tmeit — Tam) = /MD
where 1 is the length of the wires, Trit is the melt-
ing point of the wire material, an / 1 and R 1 are
the current and resistance of the first wire.
The resistance of the second wire is R2 =
R1116. Therefore, the current / 2 required for melting
the second wire must satisfy the relation
11R 2 > (^) (Tmeit — Tam).
Finally, we obtain
/2 > 811 = 80 A.
3.32. Let the emf of the second source be W2.Then,
by hypothesis,

R+R, R '

where R is the resistance of the varying resistor

for a constant current. Hence we obtain the answer:

R x = 1

3.33. The current through the circuit before the

source of emf 1 2 is short-circuited satisfies the

condition

gi

/1—

'

where r 1 and r 2 are the internal resistances of the
current sources. After the short-circuiting of the
second source of emf Wa, the current through the
resistor of resistance R can be determined from the
formula

R-1-ri •