Download Problem 2.11 Select the value of R in the circuit of Fig. P2.11 so that

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Problem 2.11 Select the value of R in the circuit of Fig. P2.11 so that VL = 9 V.
12 V
_
I0
R
+
3I0
500 Ω
6 mA
_ VL +
500 Ω
Figure P2.11: Circuit of Problem 2.11.
Solution: The voltage across the 500-Ω resistor in the right-hand segment is
VL = (3I0 + 6 × 10−3 ) × 500.
Setting VL = 9 V leads to
1
I0 =
3
VL
− 6 × 10−3
500
1
=
3
9
− 6 × 10−3
500
= 4 mA.
The left-hand loop has to satisfy KVL:
−12 + I0 R + 500I0 = 0,
which leads to
R=
12 − 500I0 12
=
− 500
I0
I0
12
=
− 500 = 2500 Ω.
4 × 10−3
c
All rights reserved. Do not reproduce or distribute. 2013
National Technology and Science Press
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