written 7.9 years ago by
teamques10
★ 65k
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modified 7.9 years ago
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1. Determining $V_{th}$ and $R_{th}$
Fig1 (a) determining $R_{th}$ (b) determining $V_{th}$
From the Fig1,
$R_{th} =R_1||R_2$
$R_{th}=35k||20k$
$R_{th}=12.72kΩ$
$I=\frac{2-(-5)}{R_1+R_2 }=\frac{7}{55k}=0.127mA$
$V_{th}-(-5V) =I R_2$
$V_{th} = (0.127m×20k)-5V$
$V_{th}= -2.46V$
The reduced circuit is as shown in fig2.
Fig2 Thevenin’s equivalent circuit
2. Calculation for ICQ
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