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

276 Aptitude Test Problems in Physics

The voltage across each of the resistors RI and

R, can be determined from the formula

gRin,

LI R2=1S-Ir-

rRs+ R 2 (r+ Rs)

The power liberated in the resistor R, can be cal-
culated from the formula

N wR,R, 12

Ra R2 L rR + Rs (Rs+ r) j R2

W 2 Rt

(rRI) 2 /R 2 d- 2rR 1 (r+ RD+ (r+ R 1 ) 2 112 •

The maximum power will correspond to the mini-

mum value of the denominator. Using the classical

inequality a 2 b 2 > tab, we find that for R, =

rRi /(r /1^1 ) the denominator of the fraction has

the minimum value, and the power liberated in

the resistor 112 attains the maximum value.

3.45. At the moment when the current through

the resistor attains the value / 0 , the charge on the

capacitor of capacitance C 1 is

C 1 / 0 R.

The energy stored in the capacitor by this moment is

W 1 — 2C1

After disconnecting the key, at the end of re-

charging, the total charge on the capacitors is q 1 ,

and the voltages across the plates of the two capac-

itors are equal. Let us write these conditions in

the form of the following two equations:

„.;

qi:F 4; - = qi, * 1 - = c v 2 ,

where and q, are the charges on the capacitors

after recharging. This gives

q qiCs

q; —

Cs

i+Cs

Cs

► Cs+ Cs •