- #1
Empire
- 3
- 0
Assuming there are five identical light bulbs. Light bulb C and D are in a series connection. C and D are in a parallel connection with E. Light Bulb A and B are in a parallel connection. The two parallel connections are wired in a series connection. What would be their relative brightness? I'm thinking that A=B=E=C+D
********************
|----power---------------|
|*******************|
|*******************|
|**|--A--|***|-C-D-|**|
|---|****|----|****|---|
***|--B--|***|--E--|***
********************
in a parallel connection, It=I1+I2, therefore, A=B. In a seriese connection, The total current equals the individual currents, so C=D. Back to the parallel connection, C+D=E. In a series, they are all equal, so (A+B)=(E+C+D). Therefore...A=B=E=C+D. My friend said that it might be: C^D<A^B<E
Any suggestions?
********************
|----power---------------|
|*******************|
|*******************|
|**|--A--|***|-C-D-|**|
|---|****|----|****|---|
***|--B--|***|--E--|***
********************
in a parallel connection, It=I1+I2, therefore, A=B. In a seriese connection, The total current equals the individual currents, so C=D. Back to the parallel connection, C+D=E. In a series, they are all equal, so (A+B)=(E+C+D). Therefore...A=B=E=C+D. My friend said that it might be: C^D<A^B<E
Any suggestions?