Two light bulbs have resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120-V line. A. Find the current through each bulb. B. Find the power dissipated in each bulb. C. Find the total power dissipated in both bulbs.

a.) O.3A and 0.15A

b.) P1 = 36W, P2 = 18W

c.) 54W

Explanation:

Constant resistance R1 = 400 ohms, R2 = 800 ohms

a.)

The current through each bulb

While in parallel.

I = 120/800/3

= 0.45A

For 400= 800/400+800

= 800/1200 x 0.45

= 0.3A

I-800 = 400 /400+800

= 400/1200*0.45

= 0.15A

B.)

The power dissipated in each bulb

P409 = I²400r

= 0.3A²x400 ohms

= 36 W

P800 = I²800R

= 0.15A²*800 ohms

= 18W

C.)

The total power dissipated

= 36W + 18 W

= 54W

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j