Question:

Please give me the below content same as it is without any changes in rendered equations **Step 1 — identify nodes and simple op-amp configurations** * Call the output of the **first** op-amp (the node to the right of (R_2), left of (R_3)) (v_{A}). The first op-amp has its noninverting ((+)) input grounded, its inverting ((-)) input connected to the junction of (R_1) and (R_2). So it is a standard inverting amplifier with input resistor (R_1) and feedback resistor (R_2). Thus [ v_A = v_{o1} = -\frac{R_2}{R_1},v_s. ] * The second op-amp has its noninverting ((+)) input grounded, its inverting ((-)) input at the junction of (R_3) (from (v_A)) and (R_4) (feedback from output). So the node at the inverting input is a virtual ground ((v=0)). Call the second op-amp output (v_{o2}). The second op-amp is an inverting amplifier with input resistor (R_3) and feedback resistor (R_4). So [ v_{o2}= -\frac{R_4}{R_3},v_{\text{in2}} ] where (v_{\text{in2}}) is the voltage applied to (R_3). Here (v_{\text{in2}}=v_A). Therefore [ v_{o2} = -\frac{R_4}{R_3},v_A. ] * The resistor (R_5) is simply between the second op-amp output and ground, so the circuit output (v_o) (voltage across (R_5), with + at the top) equals the second op-amp output: [ v_o = v_{o2}. ] **Step 2 — combine expressions** Substitute (v_A) into the expression for (v_{o2}): [ v_o = v_{o2} = -\frac{R_4}{R_3}\left(-\frac{R_2}{R_1}v_s\right) = \frac{R_2,R_4}{R_1,R_3};v_s. ] So the final expression for the output voltage is [ \boxed{,v_o = \dfrac{R_2,R_4}{R_1,R_3};v_s,.} ] **Step 3 — find (i_o)** The diagram labels a current (i_o) on the connection between the first op-amp output and the second stage (the branch that contains (R_3)). The current through (R_3) (from the left node (v_A) toward the right node at the second op-amp input) is [ i_{R_3}=\frac{v_A - v_{\text{in2}}}{R_3}. ] The second op-amp inverting input is at virtual ground, so (v_{\text{in2}}=0). Therefore [ i_{R_3}=\frac{v_A-0}{R_3}=\frac{v_A}{R_3}. ] Substitute (v_A=-\dfrac{R_2}{R_1}v_s): [ \boxed{,i_o = i_{R_3} = \frac{v_A}{R_3} = -\frac{R_2}{R_1,R_3};v_s,.} ] (That negative sign shows the actual direction of current is opposite a positive-defined direction from left→right; if the arrow on the diagram points left, use the sign accordingly. The magnitude is (|i_o|=\dfrac{R_2}{R_1R_3}|v_s|).) **Optional: current through the load (R_5)** If you also want the load current through (R_5), [ i_{R_5}=\frac{v_o}{R_5}=\frac{R_2R_4}{R_1R_3R_5};v_s. ] --- Summary (boxed): [ \boxed{,v_o=\dfrac{R_2R_4}{R_1R_3},v_s,\qquad i_o=-\dfrac{R_2}{R_1R_3},v_s,.} ]