The op-amp circuit in the figure given below is ideal

The current $I_{L}$ throug the load resistor $R_{L}$ is

\frac{- V_{s}}{R_{L}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{- V_{s}}{R_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- V_{s}}{R_{L}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- V_{s}}{R_{L}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{- V_{s}}{R_{L}}$
Redrawing the circuit

By applying KCL at node (b), we get
$\frac{V_{b}}{R_{2}} + \frac{V_{b}}{R_{L}} = \frac{V_{0} - V_{b}}{R_{2}}$

$\frac{2 V_{b}}{R_{2}} + \frac{V_{b}}{R_{L}} = \frac{V_{0}}{R_{2}}$
By applying KCL at node (a), we get
$\frac{V_{a} - V_{s}}{R_{1}} + \frac{V_{a} - V_{0}}{R_{1}} = 0$
$2 V_{a} - V_{s} = V_{0}$
$V_{a} = V_{b} = V$
$\frac{2 V}{R_{2}} + \frac{V}{R_{L}} = \frac{2 V - V_{s}}{R_{2}}$
$\frac{2 V}{R_{2}} = \frac{- V_{s}}{R_{2}}$
$\frac{V}{R_{L}} = I_{L} = \frac{- V_{s}}{R_{2}}$

Suggest Corrections

Similar questions

Z_{21}, Z_{22}

5

1

R_{i n}

k Ω

R_{L}

Join BYJU'S Learning Program