Bukhovtsev-et-al-Problems-in-Elementary-Physics

340 ANSWERS AND SOLUTIONS
/16'4
D 2 7 10
tfyolts
Fig. 453
The negative value of Ua is discarded because it does not correspond to
the sense of the problem. The sought currents are:
i1
1
(Bt+B2)R(Bui?+(A.B2-A2BdRUa-B1Ua]=22.2 rnA
i2=(B. +I B2)
R [BtB+(AtB.-A.Bt) RUa-B2Ua!=37.8 rnA
541 With the potential of the grid cC2=- 6 V, the current flowing
through the valve is 1 2 = ~ and with 111=-3 V it is 1.= ~l •
Therefore, an increase in the potential of the grid by cfJl -cfJ2=3 V raises
the anode current of the valve by
1
11-J"=R (VI -U 2)= 3.5 rnA
Since the grid characteristic of the valve in the region being considered
is assumed to be linear, the additional increase in the potential of the grid
relative to the cathode by 3 V (from-3 V to zero· with the short-circuited
gri d and cathode) will
increase the anode current by another 3.5 rnA.
The voltage drop across the resistance R will now increase additionally
by U t-Ut.=35 V, and become equal to Uo=U. +(U 1-U2 )= 130 V. while
the potential difference between the anode and the cathode of the valve will
be equal to tB-Uo=120 V. .
542. The first diode begins to conduct current only when U a > 0, I, e.,
when V > cfJlt the second at V > li". and the third at V >