الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
Solution
(a) The time constant = RC = (1×10 -6 )(5×10 -6 ) = 5sec
The charge on the capacitor = Q = Cξ = (5×10 -6 )(30) = 150 µC
(b) The current in charging of the capacitor is given by
I
e
R
−t
RC
I
= ζ 10 ⎞
⎟
30 ⎜ ⎛
−
6 −6
(1 10 )(5 10 )
⎝ × × ⎠
−6
= e
= 4.06×
10 A
6
1×
10
Example 8.13
Determine the potential difference V b -V a for the circuit shown in figure 8.22
12V
1
2 Ω
Solution
10Ω
The current is zero in the
middle branch since
there is discontinuity at 5Ω
the points a and b.
Applying the second
Kirchhoff’s rule for the
outside loop we get,
+12 - 10I - 5I - 8 - 1/2I - 1/2I = 0
+4 - 16I = 0
I = 4A
The potential difference V b -V a is found by applying the second Kirchhoff’s
rule at point a and move across the upper branch to reach point b we get,
V b -V a = +10I - 12 + 1/2I + 4
V b -V a = - 8 + 10.5I = - 8 + (10.5×4) = 34volt
a
b
4V
8V
Figure 8.22
1
2 Ω
1
2 Ω
Dr. Hazem Falah Sakeek