Solution
- R= Vg/Ig = 1000/0.005 = 200 kΩ
Pr = Ig^2 R = (.005)^2 (200,000)= 5 W
Circuit simplification by combining resistors
Resistors in Series
Resistors in series can be replaced by an equivalent resistor that is the sum of all the
resistors in series.
Resistors in Parallel
Resistors in parallel can be replaced by an equivalent resistor as shown below:
Example
Find values of I 1 and I 2 for the following circuit:
Solution:
1) Simplify the circuit by combining the two 1 M Ω parallel resistors with the 1 kΩ resistor that is in
1 kΩ (^) 1 M Ω
+
+5V
-
I 1 I 2
1 kΩ
1 M Ω
R 1 R 2... Rn^ 1/Req = 1/R 1 + 1/R 2 + ... + 1/R n
R 1 R 2 Rn
...
Req = R 1 + R 2 + ...+ R n
+
-
Vg (^) R
Ig