Calculate the output of the op-amp circuit shown in Fig.P. 18.4. Answer ( 0)




Given:

Let

Resistance 1 = `\R_1` = 10 kΩ = 10ｘ10³ Ω

Resistance 2 = `\R_2` = 4 kΩ = 4ｘ10³ Ω

Resistance 4 = `\R_4` = 20 kΩ = 20ｘ10³ Ω

Voltage across `\R_1` = `\V_1` = 5 V

Voltage across `\R_2` = `\V_2` = -2 V


To Find:

Output Voltage across `\R_3` = `\V_0` =  ?



Solution:


We will solve the question by using Kirchhoff's current rule.

From the figure 

`\I_3` =  `\I_1`  + `\I_2`   -----------(1)

So we have to find first all the three currents (`\I_1`, `\I_3`  and `\I_3` passing through the given three resistances (`\R_1`, `\R_2` and `\R_3`  respectively. So,

(1) Current `\I_1` through resistance `\R_1`:

Using Ohm's law

 `\I_1`  = `\frac {\V_1}{\R_1}`

By putting values 

 `\I_1`  = `\frac {5 V}{10ｘ10³ Ω}`

 `\I_1`  = 0.5ｘ10⁻³ A 


(2) Current `\I_2` through resistance `\R_2`:

Using Ohm's law

 `\I_2`  = `\frac {\V_2}{\R_2}`

By putting values 

 `\I_2`  = `\frac {-2 V}{4ｘ10³ Ω}`

 `\I_2`  = - 0.5ｘ10⁻³ A 


(3) Current `\I_3` through resistance `\R_3`:

Using Ohm's law

 `\I_3`  = `\frac {\V_0}{\R_3}`

By putting values 

 `\I_3`  = `\frac {V_0}{20ｘ10³ Ω}`

 `\I_3`  =  0.5ｘ10⁻³ `\V_0` A 

Now putting values in equation (1)


0.5ｘ10⁻³ `\V_0` A  = 0.5ｘ10⁻³ A + (- 0.5ｘ10⁻³ A )

0.5ｘ10⁻³ `\V_0` A  = 0 A

Or

`\V_0` = 0 V -------------------Ans.


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