Question

LAST TWO DIGIT OF ID IS 12 (b) Compute $I_B$, $I_C$, $I_E$, $V_B$ and $V_C$ for the network of Fig.Q2(b). Here, $V_{CC}$= Last two digits of your student ID+5 and $\alpha$=0.95. + $V_{CC}$ 8.2k ohm 220k ohm C2 $I_B$ $V_o$ 1 uF 330k ohm C $I_C$ Q1 C1 Beta=150 $V_i$ B 1 uF $I_B$ $\Sigma$ 8.2k ohm $I_E$ 0 Fig.Q2(b)

          LAST TWO DIGIT OF ID IS 12
(b) Compute $I_B$, $I_C$, $I_E$, $V_B$ and $V_C$ for the network of Fig.Q2(b). Here, $V_{CC}$= Last two digits of your
student ID+5 and $\alpha$=0.95.
+ $V_{CC}$
8.2k ohm
220k ohm
C2
$I_B$
$V_o$
1 uF
330k ohm
C
$I_C$
Q1
C1
Beta=150
$V_i$
B
1 uF
$I_B$
$\Sigma$
8.2k ohm
$I_E$
0
Fig.Q2(b)

LAST TWO DIGIT OF ID IS 12
(b) Compute IB, IC, IE, VB and VC for the network of Fig.Q2(b). Here, VCC= Last two digits of your
student ID+5 and α=0.95.
+ VCC
8.2k ohm
220k ohm
C2
IB
Vo
1 uF
330k ohm
C
IC
Q1
C1
Beta=150
Vi
B
1 uF
IB
Σ
8.2k ohm
IE
0
Fig.Q2(b)

Added by Elizabeth F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Rajesh Kumar

Step 1

Vcc = Last two digits of your student ID + 5 = 12 + 5 = 17. Show more…

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Please give me my answer soon. Then I will vote you. SUBJECT NAME: Electronic devices and circuits , EEE [NB: Please dear, give me carefully correct answer] Question: LASTTWO DIGITOF ID IS 12 (b) Compute I, Ic,IE,V and Vc for the network of Fig.Q2(b. Here, Vcc=Last two digits of your student II+5 and =0.95 +Vcc 8.2k ohm 220k ohm C 2 18 W 1uF 330k ohm c Ic Q1 C1 B 8.2k ohm Fig.Q2(b)