Express the voltages v1 , v2n the circuit of Fig. 3.27(a) in terms of the source voltage v0 where R1 = 1 kΩ , R2 = 2.2 KΩ , and R3 = 10 KΩ .

Question

Express the voltages [latex] u_1 , u_2 [/latex] in the circuit of Fig. 3.27 (a) in terms of the source voltages [latex] u_0 [/latex], where [latex] R_1=1 k \Omega, R_2=2.2 k \Omega [/latex], and [latex] R_3= 10 k \Omega [/latex].

Accepted Answer

We identify and label the nodes as shown in Fig. 3.27(b), choosing node d as the reference node. By the definition of an ideal voltage source, [latex] u_a= u_0 [/latex] . The node voltages [latex] u_b= u_{b d} [/latex] and [latex] u_c= u_{c d} [/latex] are unknown, so we apply Kirchhoff's current law to those nodes. Thus

[latex] \begin{aligned} i_1+i_2+i_3 &=\frac{ u_b- u_0}{R_1}+\frac{ u_b}{R_2}+\frac{ u_b- u_c}{R_1}=0 \\ & \Rightarrow\left(\frac{2}{R_1}+\frac{1}{R_2} ight) u_b-\frac{ u_c}{R_1}=\frac{ u_0}{R_1}, \\ -i_3+i_4+i_5 &=\frac{ u_c- u_b}{R_1}+\frac{ u_c}{R_2}+\frac{ u_c- u_0}{R_3}=0 \\ & \Rightarrow-\frac{ u_b}{R_1}+\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} ight) u_c=\frac{ u_0}{R_3} . \end{aligned} [/latex]

To simplify the notation, we use conductances in place of resistances; e.g., [latex] G_1=1 / R_1 [/latex] .

Thus

[latex] \begin{aligned} &\left(\frac{2}{R_1}+\frac{1}{R_2} ight) u_b-\frac{ u_c}{R_1}=\frac{ u_0}{R_1} \\ &\Rightarrow\left(2 G_1+G_2 ight) u_b-G_1 u_c=G_1 u_0 \\ &\quad -\frac{ u_b}{R_1}+\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} ight) u_c=\frac{ u_0}{R_3} \\ &\Rightarrow-G_1 u_b+\left(G_1+G_2+G_3 ight) u_c=G_3 u_0 . \end{aligned} [/latex]

To simplify further, let

[latex] \begin{aligned} G_b &=\frac{2}{R_1}+\frac{1}{R_2} \cong 2.455 mS \\ G_c &=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \cong 1.555 mS . \end{aligned} [/latex]

whence

[latex] G_b u_b-G_1 u_c=G_1 u_0,-G_1 u_b+G_c u_c=G_3 u_0, [/latex]

which yield

[latex] \begin{aligned} & u_b= u_1=\frac{G_1\left(G_c+G_3 ight)}{G_b G_c-G_1^2} u_0 \cong 0.588 u_0, \\ & u_c= u_2=\frac{G_b G_3+G_1^2}{G_b G_c-G_1^2} u_0 \cong 0.442 u_0 \end{aligned}. [/latex]

Question 3.8: Express the voltages v1 , v2n the circuit of Fig. 3.27(a) in...

Related Answered Questions

Using superposition, obtain an expression for the current ix in the circuit shown in Fig. 3.44(a). ...

Verified Answer:

Refer to Fig. 3.29(a). Express the voltage ν0 in terms of the circuit parameters. ...

Verified Answer:

Refer to Fig. 3.28(a). Express the voltage ν2 in terms of the circuit parameters. ...

Verified Answer:

Refer to Fig. 3.38. Use mesh analysis to obtain an expression for the voltages ν1 , ν2 , ν3 . ...

Verified Answer:

Obtain an expression for the current i1 through the resistor R1in the circuit of Fig. 3.37. ...

Verified Answer:

(Compare with Example 3.6): Obtain an expression for the voltage across the resistor R2 in the circuit of Fig. 3.35. The resistances and source voltage are presumed to be known. ...

Verified Answer:

Obtain an expression for the current i in the circuit of Fig. 3.36. The resistances and source voltages are presumed to be known. ...

Verified Answer:

In Fig. 3.32, ν1 =3 V, ν2 = 10 V, ν4 = 15 V Find the voltage ν3. ...

Verified Answer:

In the circuit of Fig. 3.16, the source voltages are ν1 = 5 V and ν2 = 15 V across the (unspecified) circuit element. Find the voltage νx ...

Verified Answer:

Obtain an expression for the current i2 in the circuit of Fig. 3.26(a). The resistances and the source voltages are presumed to be known. ...

Verified Answer: