I have trouble on physics question, please answer it and show the work. thank yo

Question

I have trouble on physics question, please answer it and show the work. thank you so much

In the figure below, you will attach one or more light bulbs of R=36.12 Ohm in parallel with the battery (with epsilon = 14 V, internal resistance r). For an ideal battery (r=0), what would be power dissipated in the first bulb if it was: connected alone: connected as one of four bulbs: (How do these compare?) For a non-ideal battery where r=2.1 H, Re-answer part a. Also, in each case find the voltage drop that would be measured across the battery terminals: connected alone: connected as one of four bulbs: P = W (Compare your P values to part a. Compare your V_batt values to epsilon. Explain.) Suppose you wish to dim the first bulb, such that its power output is at or below __ % of its value when connected alone. Find the minimum number of total bulbs you must connect in parallel to make this happen. Solve for a general expression so you can use it for all the cases below. 50%: bulbs 25%: bulbs 15%: bulbs

Explanation / Answer

From the given question,

Resistance of bulb(R)=36.12 ohms

Emf of battery(E)=14V

Internal resistance(r)=r

(a) for ideal battery when r=0,

power dissipated= E2/R= 142/36.12=5.426 W

four bulbs parallel:

Since it is a parallel circuit, the emf is same for all bulbs.

so all the bulbs will dissipate same power.(5.426W)

(b)For non ideal battery r=2.1ohm

I= E/(R+r)

=14/(36.12 + 2.1)=0.366amp

Terminal voltage= E-Ir=14- 0.366 x 2.1=13.23V

Power of bulb = V2/R=(13.23)2/36.12=4.845W

four bulbs parallel

Net resistance of all four bulbs= 36.12/4=9.03

I= 14/(9.03 + 2.1)=1.258A

Terminal voltage= E-Ir=14- (1.258)(2.1)=11.35V

Power dissipated= V2/R= (11.35)2/ 36.12=3.566W

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---