Parallel example
Three resistors, with values of 8 W, 8 W,
and 4 W are connected in parallel with one another and with a 10-volt battery.
(a) What is the total current provided by the battery?
(b) What is the power dissipated in each resistor?
First find the equivalent resistance, which is:
| 1
|  |
| Req
|
|
= |
| 1
| |
| R1
|
|
+ |
| 1
| |
| R2
|
|
+ |
| 1
| |
| R3
|
|
= |
| 1
| |
| 8
|
|
+ |
| 1
| |
| 8
|
|
+ |
| 1
| |
| 4
|
|
= |
| 4
| |
| 8
|
|
= |
| 1
| |
| 2
|
|
Flip this upside down to get Req = 2 W
| I |
= |
| DV
| |
| Req
|
|
= |
| 10
| |
| 2
|
|
= |
5 A |
The current through each resistor can be found using Ohm's Law.
| For each 8 W resistor, I |
= |
| DV
| |
| R
|
|
= |
| 10
| |
| 8
|
|
= |
1.25 A |
| For the 4 W resistor, I |
= |
| DV
| |
| R
|
|
= |
| 10
| |
| 4
|
|
= |
2.5 A |
The sum of the currents equals the total current from the battery, as it should.
The power dissipated in each resistor can be found a number of different ways. Here's one method:
| For each 8 W resistor, I |
= |
| DV2
| |
| R
|
|
= |
| 10*10
| |
| 8
|
|
= |
12.5 W |
| For the 4 W resistor, I |
= |
| DV2
| |
| R
|
|
= |
| 10*10
| |
| 4
|
|
= |
25 W |
That's a total of 50 W. Check this against the power provided to the circuit by the battery:
P = DV I = 10 * 5 = 50 W.
They agree, as they should.