The circuit shows a source of EMF and some identical light bulbs. The switch, S,

Question

The circuit shows a source of EMF and some identical light bulbs. The switch, S, is opened at some point. The questions refer to the situation before and after the opening of S. Answer true or false for each statement.
(True/False) When S is opened, bulb 2 becomes brighter.
(True/False)  When S is closed, bulb 1 is brighter than bulb 4.
(True/False)  When S is opened, the current through bulb 2 decreases.
(True/False)   When S is opened, the potential differences across bulbs 1 and 4 are unaffected.
(True/False)   When S is opened, the current through bulb 3 is zero.

The circuit shows a source of EMF and some identical light bulbs. The switch, S, is opened at some point. The questions refer to the situation before and after the opening of S. Answer true or false for each statement. (True/False) When S is opened, bulb 2 becomes brighter. (True/False) When S is closed, bulb 1 is brighter than bulb 4. (True/False) When S is opened, the current through bulb 2 decreases. (True/False) When S is opened, the potential differences across bulbs 1 and 4 are unaffected. (True/False) When S is opened, the current through bulb 3 is zero.

Explanation / Answer

true

true

false

false

true

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explanation

consider each bulb of resitance r

intially S is closed.

total resistance = r/3 + r = 4r/3

net current through battery = 3V/4r = I

current through 1 is I

current through 2 , 3 and 4 is I/3

voltage across 1 is 3V/4r * r = 3V/4

voltage across 2,3 and 4 is V/4r * r = V/4

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after switch is opened.

net resistance = r + r/2 = 3r/2

net current = 2V/3r

current through 1 is 2V/3r
current through 2 and 4 is V/3r

current through 2 is 0

voltage across 1 is 2V/3

voltage across 2,3 and 4 is V/3

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#1

when S is opened , bulb 2 becomes brighter

before opening voltage across bulb 2 is V/4 and current is V/4r

after opening voltage across bulb 2 is V/3 and currenr is V/3r

since brightness is nothing but power dissipation

brightness before opening is VI = V^2/16r

brightness after opening is VI = v^2/9r

there after opening brightness improved so statement is correct so

true

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#2

when S is closed , bulb 1 is brighter than bulb 4

when closed voltage across bulb 1 is 3V/4 and current is 3V/4r

that across bulb 4 is V/4 and V/4r

since current and voltage across 1 is more.

brightness of 1 is more than 4.

so true

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#3

When S is opened, the current through bulb 2 decreases

before opening current through 2 is V/4r

after opening current through 2 is V/3r

actually current through 2 is increased when S is opened

so statement is False

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#4 When S is opened, the potential differences across bulbs 1 and 4 are unaffected.

across 1 potential difference before opening is 3V/4 and after opening is 2V/3

that across 4 is V/4 before opening and V/3 after opening

so they are not same.

so the statement is false.

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#5 When S is opened, the current through bulb 3 is zero.

yes when S is opened , 3 has become open circuit

so no current will flow. hence zero current

so the statement is True.

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