Question

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5. The non-series-parallel circuit to the right is known as a twin-T network Suppose that R2=27Ω and v2=1 V. Find Ri and the single equivalent resistance Req Ri 5Ω 2Ω

Answer 1

Ans)

The circuit labeled with voltages and currents is

The current through R₂ by ohms law is

U21

Current through 2 ohms resistor is

$I_{2\Omega }=\frac{v-v_{2}}{2}=\frac{9-1}{2}=4\: \: A$

$I_{2\Omega }=4\: \: A$

Using Kirchoff's current law

$I_{1\Omega }=I_{2\Omega }-I_{R_{2}}=4-3.5=0.5\: \: A$

$I_{1\Omega }=0.5\: \: A$

Using Kirchoff's voltage law

$v_{3}=v_{2}-I_{1\Omega }*1=1-0.5*1=0.5\: \: V$

$I_{\frac{1}{3}\Omega }=\frac{v_{3}}{\frac{1}{3}}=\frac{0.5}{\frac{1}{3}}=1.5\: \: A$

$I_{\frac{1}{3}\Omega }=1.5\: \: A$

$I_{5\Omega }=I_{\frac{1}{3}\Omega }-I_{1}=1.5-0.5=1\: \: A$ KCL

$I_{5\Omega }=1\: \: A$

$v_{1}=v_{3}+I_{5\Omega }*5=0.5+1*5=5.5\: \: V$

$I_{3\Omega }=\frac{v_{1}}{3}=\frac{5.5}{3}=1.833\: \: A$

$I_{3\Omega }=1.833\: \: A$

$I_{R_{1}}=I_{3\Omega }+I_{5\Omega }=1.833+1=2.833\: \: A$

The voltage across resistor R1 by KVL is

$R_{1}=\frac{V_{R_{1}}}{I_{R_{1}}}=\frac{3.5}{2.833}=1.235\: \: \Omega$

$\mathbf{R_{1}=1.235\: \: \Omega }$

By KCL

$i=I_{R_{1}}+I_{2\Omega }=2.833+4=6.833\: \: A$

Now we will find single equivalent resistance as below

$R_{eq}=\frac{v}{i}=\frac{9}{6.833}=1.317\: \: \Omega$

$\mathbf{R_{eq}=1.317\: \: \Omega }$