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?
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
- PearlsawmeLv 78 years agoFavorite 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 .
- ColinLv 78 years ago
I assume that R is the resistance of a bulb.
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) <<<
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.
- akm69Lv 78 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
- Anonymous6 years ago
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