Solutions (^271)
The voltages U and U (^3) across the capacitors can
be calculated from the formulas
U1—
R2C2
Cl (r R 2 ) (CI+ CO
CB (r+ R (^2) ) (C1- 1 - C2)
respectively.
3.38*. We shall mentally connect in series two per-
fect (having zero internal resistance) current sources
of emf's equal to — Us. and U 0 between points A
and F. Obviously, this will not introduce any
change in the circuit. The dependence of the current
through the resistor of resistance R on the emf's of
the sources will have the form
I = — v. ± 01/ 0 ,
where (^) is the emf of the source contained in the
circuit, and the coefficients a and 13 depend on the
resistance of the circuit.
If we connect only one perfect source of emf
equal to — U 0 between A and F, the potential
difference between A and B will become zero.
Therefore, the first two terms in the equation
for I will be compensated: I = pt.1 0. The coefficient
13 is obviously equal to 1/(R + Reff ), where R e ft
is the resistance between A and /13 when the re-
sistor R is disconnected. This formula is also valid
for the case R = 0, which corresponds to the con-
nection of the ammeter. In this case,
Io „
fleff
Consequently, the required current is
I U 0 0 /
RI0 + U 0
3.39. When the key K is closed, the voltage
across the capacitor is maintained constant and
equal to the emf of the battery. Let the
displacement of the plate B upon the attainment
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