الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Direct Current Circuits
+ξ 1 - IR 1 - ξ 2 + IR 2 = 0
Solving for the current we get
ζ
1
−ξ
2
I =
R + R
1
2
6 −12
1
= = − A
8 + 10 3
The -ve sign of the current indicates that the correct direction of the current
is opposite the assumed direction i.e. along the loop (adba)
The power dissipated in R 1 and R 2 is
P 1 = I 2 R 1 = 8/9W
P 2 = I 2 R 2 = 10/9W
In this example the battery ξ 2 is being charged by the battery ξ 1 .
Example 8.4
Three resistors are connected in series with battery as shown in figure
8.7, apply second Kirchhoff’s rule to (a) Find the equivalent resistance
and (b) find the potential difference between the points a and b.
a
R 1
ξ
I
I
I
R 2
b
R 3
Figure 8.7
Solution
Applying second Kirchhoff’s rule in clockwise direction we get
- IR 1 - IR 2 - IR 3 + ξ = 0
or
I =
R
ζ
1
+ R2
+ R3
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