FRAC{{{V_b}}}{{100\;K}} + \frac{{{V_b} - {V_2}}}{{10\;k}} = 0\) \(\RIGHTARROW \frac{{{V_b}}}{{100\;k}} + \frac{{{V_b} - 50\;m}}{{10\;k}} = 0\) \(\Rightarrow {V_b} + 10\;{V_b} - 500\;m = 0 \Rightarrow {V_b} = \frac{{0.5}}{{11}}V\) By applying KCL at a,\(\frac{{{V_a} - {V_1}}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) \(\Rightarrow \frac{{{V_a} - 10\;m}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) ⇒ 10 Va – 100 m + Va - Vout = 0⇒ Vout = 11 Va – 0.1\({V_a} = {V_b} = \frac{{0.5}}{{11}}V\)\(\Rightarrow {V_{out}} = 11\left( {\frac{{0.5}}{{11}}} \right) - 0.1 = 0.4 = 400\;mV\)

"> FRAC{{{V_b}}}{{100\;K}} + \frac{{{V_b} - {V_2}}}{{10\;k}} = 0\) \(\RIGHTARROW \frac{{{V_b}}}{{100\;k}} + \frac{{{V_b} - 50\;m}}{{10\;k}} = 0\) \(\Rightarrow {V_b} + 10\;{V_b} - 500\;m = 0 \Rightarrow {V_b} = \frac{{0.5}}{{11}}V\) By applying KCL at a,\(\frac{{{V_a} - {V_1}}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) \(\Rightarrow \frac{{{V_a} - 10\;m}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) ⇒ 10 Va – 100 m + Va - Vout = 0⇒ Vout = 11 Va – 0.1\({V_a} = {V_b} = \frac{{0.5}}{{11}}V\)\(\Rightarrow {V_{out}} = 11\left( {\frac{{0.5}}{{11}}} \right) - 0.1 = 0.4 = 400\;mV\)

">
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In the circuit below, the operational amplifier is ideal. If V1 = 10 mV and V2 = 50 mV, the output voltage (Vout) is

Electronics & Communication Engineering Op-Amp And Its Applications in Electronics & Communication Engineering . 7 months ago

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By applying KCL at Vb,\(\FRAC{{{V_b}}}{{100\;K}} + \frac{{{V_b} - {V_2}}}{{10\;k}} = 0\) \(\RIGHTARROW \frac{{{V_b}}}{{100\;k}} + \frac{{{V_b} - 50\;m}}{{10\;k}} = 0\) \(\Rightarrow {V_b} + 10\;{V_b} - 500\;m = 0 \Rightarrow {V_b} = \frac{{0.5}}{{11}}V\) By applying KCL at a,\(\frac{{{V_a} - {V_1}}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) \(\Rightarrow \frac{{{V_a} - 10\;m}}{{10\;k}} + \frac{{{V_a} - {V_{out}}}}{{100\;k}} = 0\) ⇒ 10 Va – 100 m + Va - Vout = 0⇒ Vout = 11 Va – 0.1\({V_a} = {V_b} = \frac{{0.5}}{{11}}V\)\(\Rightarrow {V_{out}} = 11\left( {\frac{{0.5}}{{11}}} \right) - 0.1 = 0.4 = 400\;mV\)

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Posted on 10 Nov 2024, this text provides information on Electronics & Communication Engineering related to Op-Amp And Its Applications in Electronics & Communication Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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