5. The figure shows two light bulbs connected to a battery of emf 1. The power d
ID: 1788508 • Letter: 5
Question
5. The figure shows two light bulbs connected to a battery of emf 1. The power delivered to these light bulbs is 30 W and 60 W as shown. Which statement about the currents lf and ld through the light bulbs is correct? 60 1 E. No conclusion about the currents is possible without the values of the resistances of the light bulbs. 6. The electric potential in a region between x-0 and x -6.00 m is a hr,where a 12.0 V and-5.00 Vi The force exerted on a charge q-0.60 C placed at -0.80 m is A -169 x 10N B -3.06x 10*N C. -2.10×10-3 N D. +4.8×104 N E. +12x 10 N 7. The switch S in the figure below is closed for a period of time that is long enough to ensure that the capacitor is fully charged I R-15.0 k2, what is the potential difference across the capacitor? A 90 V B 5.0V C. 30v D. 60V E 7.5 V 12.ou 9.00 V s. Io the cicuit shown, R-600 a C- 10andticena resisans of hebanny All five capacitors are initially uncharged. The total charge stored in the circuit after 15 s is A. 7.1 x 10 c B. 94 x 10*c C. 3.8 x 10c D. 6.5 × 10-6 c E. 2.2x 10 cExplanation / Answer
5. power of light bulbs = 30W and 60 W
potnetial difference = dV
now, Power P = dV*I
as dV is same for both the bulbs
Icd = 2*Ief
Ief = 0.5Icd
hence
option A
6. x = 0, x = 6m
V = a + bx^2
-dV/dx = -2bx
b = -5
hence
E = -dV/dx = 10x
so force on q= 0.6 micro C
at x = 0.8 m is
F = qE = 0.6 micro * 10*0.8 = 4.8 micro N
hence option D
7. for fully charged capacitor
R = 15,000 ohm
potential diff across capacitor = R*9/(12,000 + R)
V = 9*15/(12 + 15) = 5 V
option B
8. R = 600 ohm
C = 100 micro F
dV = 12 V
Ceff = 5C = 500 micro F
Reff = 5R = 3000 ohm
charging equation
q = qo*(1 - e^(-t/RC))
qo = Ceff*V = 500*10^-6*12 = 0.006 C
hence
q = 0.006(1 - e^(-1.5/500*3000*10^-6)) = 0.003792 C
hence option C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.