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5 identical bulbs with equal resistance R in a circuit, can you explain why the answers are right?

Figure 2: http://i48.tinypic.com/eamfdz.jpg

(Figure 1 is the same but with the switch closed)

I know the answers to these questions but i need to understand why.

1) With the switch open, what is the current flowing through C in terms of EMF (E) and R?

answer: 2E/3R

For this i added up the resistors together to find the total resistance and divided E by this total resistance to get I.

2) What is the current flowing in bulb C when the circuit is closed?

The total resistance I got is 7/6R. But I don't know how they used that answer to obtain I=(6/7)(2E/3R) for the current in bulb C when it is closed. Can you explain in detail how they got this?

3) What happens to bulb C when the circuit is opened?

a) It gets brighter

b) It gets dimmer

c) It's brightness stays the same.

answer: It gets brighter. This I don't understand as well

4 Answers

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  • 8 years ago
    Favorite Answer

    The equivalent resistance of A and B is R’ = R*R/( R+R) = R/2

    ( since they are in parallel )

    Hence in the figure we can replace A and B and keep one bulb of resistance R/2.

    When switch is open the current flows through R/2 and R of the bulb C.

    No current flows through D and E.

    Total resistance is R + R/2 = 3 R/ 2

    I = E /( 3R/2) = 2 E/ 3R

    Sorry to say that Your explanation

    “ For this i added up the resistors together to find the total resistance and divided E by this total resistance to get I. “ is wrong .

    ====================================

    If C is closed we have to find the total reistance of the parallel combination of the three bulbs

    The resistance of D and E is 2R and this is in parallel with R.

    Equivalent resistance is 2R*R /( 2R + R ) = 2R/ 3.

    This is in series with the already calculated R/2

    Hence it is R/2 + 2R/ 3 = (7 R / 6 )

    Current I = E / [(7R/6) = 6E /( 7R)

    This is the total current.

    But we have to find the current in the bulb C alone..

    If I c is the current through C, and the current through D and E is I de

    Equating the potentials at the ends of the resistances

    Ic R = I de * ( 2R )

    Ic = 2 I de

    But Ic + I de =6E /( 7R)

    But Ic + Ic/ 2 = 6E /( 7R)

    3 Ic / 2 = 6E /( 7R)

    Ic = (2/3) 6E /( 7R)

    ================================

    When the switch is closed the current we have found as (2/3) 6E /( 7R) =====1

    When switch is opened the current was 2E / 3R==========2

    1 / 2 gives 6/ 7 which is less than 1 .

    The current when the switch is closed is less

    Hence it is brighter when switch is opened .

    ===============================

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  • Colin
    Lv 7
    8 years ago

    I assume that R is the resistance of a bulb.

    1)

    What's the total resistance across the battery?

    The resistance of A and B in parallel is R/2.

    That must be added to the resistance of C, ie R.

    Total resistance = 3*R/2

    Current = volts/resistance = E*2/(3*R) <<<

    2)

    With the switch closed, again, what's the total resistance?

    A in parallel with B = R/2 as before.

    D + E = 2*R

    D + E in parallel with C has resistance R1 where

    1/R1 = 1/R + 1/(2*R) = 3/(2*R)

    R1 = 2*R/3

    Total resistance = (R/2) + (2*R)/3 = 7*R/6

    Total current = E*6/(7*R)...agreed

    There's the same voltage across C and (D and E in series).

    Current is inversely proportional to resistance...i=v/r

    Thus twice the current flows in R (ie bulb C) compared to 2*R (ie bulbs D + E).

    So 2/3 flows in C and 1/3 in D + E

    Hence current in C is (2/3)*E*6/(7*R) <<<

    3) In other words, is there more current in C when the switch is opened?

    Rearranging the result from 2) so that we can more easily compare it with the result from 1)

    I = E*2/(3*R)*(6/7)

    This is the current with the switch *closed*; you can now see that it is (6/7) the current with the switch *open*.

    Hence more current with switch open, and so the bulb is brighter with the switch open.

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  • akm69
    Lv 7
    8 years ago

    1) A & B are in parallel

    =>1/R(A&B) = 1/R + 1/R

    =>R(A&B) = R/2Ω

    Now R(A&B) and C are in series

    =>R(net) = R/2 + R

    =>R(net) = 3R/2Ω

    By V = iR

    =>E = i x 3R/2

    =>i = 2E/3R

    2)D&E are in series:-

    =>R(D&E) = R + R = 2RΩ

    R(D&E) and C are in parallel:-

    =>1/R(D,E&C) = 1/2R + 1/R

    =>R(D,E&C) = 2R/3Ω

    A&B are in parallel

    =>R(A&B) = R/2Ω

    Now R(A&B) and R(D,E&C) are in series:-

    =>R(net) = R/2 + 2R/3 = 7R/6Ω

    Thus by V = iR

    =>i(net) = E/(7R/6) = 6E/7R amp and this will be same across R(A&B) and R(D,E&C) as they are in series:-

    thus By V = iR in R(D,E&C)

    =>V = 6E/7R x 2R/3

    =>V(across D,E&C) = 4E/7

    Now C & (D,E) are in parallel thus V will be across both

    =>By V = iR

    =>4E/7 = i(C) x R

    =>i(C) = 4E/7R

    3) a) It gets brighter

    As in open circuit current through C is 2E/3R > than the current in close circuit i.e. 4E/7R

    & By P = i^2 R

    =>P(open)>P(close)

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  • Anonymous
    6 years ago

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