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

Question

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

Answer 1

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 .

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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)





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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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Answer 2

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.

Answer 3

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)

Answer 4

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