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}}

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

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

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

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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
$\frac{V_{a} - V_{s}}{R_{1}} + \frac{V_{a} - V_{0}}{R_{1}} = 0$
$2 V_{a} - V_{s} = V_{0}$
$V_{a} = V_{b} = V$
$\frac{2 V}{R_{2}} + \frac{V}{R_{L}} = \frac{2 V - V_{s}}{R_{2}}$
$\frac{2 V}{R_{2}} = \frac{- V_{s}}{R_{2}}$
$\frac{V}{R_{L}} = I_{L} = \frac{- V_{s}}{R_{2}}$

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