الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Current & Resistance
R 4
R 2
R 1
R 3
R 5
Figure 7.7
Solution
Resistance R 1 and R 2 are connected in series therefore the circuit is simplify
as shown below
R 4
R 5
R 3
R 1 &R 2
R 4
R 5
R 1 &R 2 &R 3
R
R′ =R 1 +R 2 =4+3=7Ω
Then the resultant resistance of R 1 &R 2 (R′ ) are connected in parallel with
resistance R 3
1 1 1 1 1 10
= + = + =
R′ R′
R 7 3 21
3
R ′ =2.1Ω
The resultant resistance R for R 5 &R 4 & R ′ are connected in series.
R= R ′ + R 5 +R 4 =2.1+5+2.9=10Ω
I
a
Example 7.8
Three resistors are connected in
parallel as in shown in figure 7.8. A
potential difference of 18V is
maintained between points a and b.
18V
I 1
I 2
I 3
R 1
R 2
R 3
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b
Figure 7.8