HoMEWORK FOR LAB 6 AND OHM\'S t below, the battery maintains i . In the circui a

Question

HoMEWORK FOR LAB 6 AND OHM'S t below, the battery maintains i . In the circui a constant potential difference the battery between its terminals, points 1 and 2 (i.e., the internal resistance of is considered negligible). The three light bulbs, A, B, and C are identical. How do the brightnesses of the three bulbs compare to each other? Explain your reasoning. a. b. What happens to the brightness of each of the three bulbs when bulb A is unscrewed and removed from its socket? Explain your reasoning. When A is unscrewed, what happens to the current through points 3, 4, and 5? Explain your reasoning. c. Bulb A is screwed back in. What happens to the brightness of each of the three bulbs when bulb C is unscrewed and removed from its socket? Explain your reasoning. d. e. When C is unscrewed, what happens to the current flowing through points 3, 4, and 5? Explain your reasoning. For each of the questions A-E below, a wire is connected from the battery ter- minal at point 1 to point 4. 2. a. What happens to the brightness of each of the three bulbs? Explain. 6: VOLTAGE IN SIMPLE DC CIRCUITS AND OHM'S LAW

Explanation / Answer

a)

Brightness of the each bulb will depend on the amount of heat energy produced by each bulb,

and brightness will be directly proportional to the amount of heat energy produced

Since each bulb are identical, so take R be the resistance of each bulb,

As we know power produced by each bulb = I2R

so, amount of heat energy produced, H = Power * time = I2Rt

since R and t will be same for each bulb , so,

Heat energy H will be directly proportional to I2

Let I be the current flowing through bulb A, then current flowing through bulb B and C will be I/2 as bulb B and C are parallel to each other and combination of B and C is in series wih bulb A,

so, Ratio of current for A,B and C is as , IA:IB:IC =  I :I/2 :I/2 = 2: 1 : 1,

therefore ratio of brightness = ratio of Heat energy (I2)= 4 : 1 : 1

so, Order of brightness will be as Bulb A > Bulb B = Bulb C

b)

When Bulb A is unscrewedso, it will go out , Nothing will change for the Bulb B and C as bulb A was parallel to the Bulb B and bulb C, so power across the bulb B and bulb C will remain same, so Bulb B and bulb C will have same brightness as in part (a)

c)

Since B and C are in series só, Combined resistance of B and C will be as RBC =  R + R = 2R

also, Since RBC is parallel to RA

só, net resistane of the circuit, Rnet = RA * RBC / (RA + RBC) = R * 2R/(R+2R)= 2R/3 ,

so, I3 = V/ (2R/3) = 3V/2R,

I4 = V/2R

and I5 = V/ R

when A is unscrewed, net resistance of the circuit will be , Rnet' = R + R = 2R,

so, I3 = V/2R

I4 = V/ 2R,

I5 = 0 (circuit is incomplete)

so current I3 will drop by factor of 1/3 , I4 will remain same and I5 = 0

d)

When bulb C is unscrewed ,

Nothing will change for bulb A as it is parallel to the circuit from where bulb is removed, so it will have same brightness as in part (a)

Since the net resistance in the branch in which bulb B is present is decreased so, the brightness of bulb B will increase and Bulb C has already removed.

e)

When, bulb C is removed, Net resistance will be = R*R/2R = R/2,

since net resistance has decreased, so current I3 will increase,

Since resistance in Bulb B branch has increased so, I4 will increase

since there will be no effect on Bulb A branch due to unscrewed of bulb C, so, current I5 will remain same.

2)

When a wire is connected between point 1 and point 4, it will short circuit complete curcuit as current always choose easy path to flow, so no current will flow through any bulb so, all bulbs will go out i.e. no bulb will glow.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.