Consider the op-amp circuit shown in the figure below:

the resistance $R_{i n}$ as seen by the voltage source is ____ $k Ω$
-100

Open in App

Solution

The correct option is A -100

now, $I_{i n} = \frac{V_{A} - V_{0}}{1 M Ω}$ ...(i)

Applying KCL at inverting terminal, we get,

$\frac{V_{B} - 0}{10 k Ω} + \frac{V_{B} - 0}{100 k Ω} = 0$

$\frac{V_{B}}{10 k Ω} = \frac{V_{0} - V_{B}}{100 k Ω}$ ... (ii)

now $V_{B} = V_{A}$ due to virtual ground

thus, $\frac{V_{A}}{10 k Ω} = - [\frac{V_{A} - V_{0}}{100 k Ω}]$

$∴ \frac{V_{A}}{10 k Ω} = \frac{- I_{i n} \times (1 m Ω)}{100 k Ω}$

$\frac{V_{i n}}{I_{i n}} = - 100 k Ω$ $(∵ V_{A} = V_{i n})$

Suggest Corrections

Similar questions

R_{i n}

k Ω

V_{0}

^{'} β^{'}

V_{s}

Ω

Join BYJU'S Learning Program