Physics, asked by joelle96, 9 months ago

Given a figure:

A-When switches S1 and S2 are open, the ammeter reading is 0.6 A and Voltmeter reading is 2.4 V calculate the resistance of the lamp and of R1

B-Compare the brightness of the lamp when both switches are open as in (a) and when only switch S1 is closed. Give a quantitative explanation.

C-Compare the brightness of the lamp when both switches are open as in (a) and when both switches S1 and S2 are closed. Give a quantitative explanation

D-For R2 = 4 Ω, if switch S1 is opened but switch S2 is closed, what is the ammeter reading? What is the voltmeter reading?


This is what i have so far:

A.

R_lamp=V/I=2.4/0.6=4Ω

R_eq=V/I=6/0.6=10Ω; R_eq=R_1+R_lamp so R_1=R_eq-R_lamp=10-4=6Ω


B.

The brightness remains the same when both switches are open and when only switch S1 is closed as nothing is really connected to the branch containing S1.


C.

When both switches S1 and S2 are open, the total resistance of the circuit decreases while the current and the dissipation of power increase. This would mean that the bulbs are brighter when the switches are closed in comparison to when they are open.


D.

Ammeter: R_1=(R_lamp×R_2)/(R_lamp+R_2 )=(4×4)/(4+4)=2Ω; R_eq=2+6=8Ω;I=V/R=6/8=0.75A

Voltmeter: V=IR=0.75×4=3V

Attachments:

Answers

Answered by AditiHegde
1

Given:

The circuit diagram.

To find:

The answers for (a), (b), (c) and (d)

Solution:

a)The voltage across lamp is:

V = 2.4 V  as  S2 is open and current flows through lamp only,

so, we have, I = 0.6

R = V/I = 2.4/0.6 = 4 Ω

Therefore, R = 4 Ω

Req = R1 + Rlamp

I = 6V/Req

Req = 6/0.6 = 10  Ω

R1 + Rlamp = 10

⇒ R1 = 10 - 4 = 6 Ω

Therefore, R1 = 6 Ω

The resistance of the lamp and of R1 are 4 Ω  and 6 Ω  respectively.

b) As there is no circuit connected to the branch containing S1, the brightness remains the same for when both switches are open as in (a) and when only switch S1 is closed.

Brightness_{open} = Brightness_{S1_closed}

c)When the switches are open, the total resistance of the circuit is less and more current is flowing in the circuit (I ∝ 1/R) and more power dissipation (P ∝ I) occurs so the bulb is brighter than as compared to when both switches are closed.

Brightness_{open} > Brightness_{S1 and S2_closed}

d) From given, we have, R2 = 4 Ω

R1 = 4 × 4/(4 + 4) = 2 Ω

Req = 2 + 6 = 8  Ω

I = 6/8

I = 0.75 A

V = IR = 0.75 × 4 = 3 V

The ammeter reading is 0.75 A and the voltameter reading is 3 V

Similar questions