Op Amp Impedance Matching

Circuit 2: For an ideal op amp, there is no current flowing through the ( $+$ ) and ( $-$ ) terminals and hence no current through the internal resistance $R_{in} = 10 \ \Omega$ and no voltage drop across $R_{in}$ . The voltage at terminal ( $+$ ) is same as $V_{in}=1 \ V$ of the DC voltage source. And for an ideal op amp, voltage at ( $+$ ) is same as that of ( $-$ ). Therefore, the output voltage across the load $1 \ \Omega$ is same as the input voltage of the DC voltage source that is $1 \ V$ . This op amp configuration is in fact called voltage follower . The load power $P_2 = \dfrac {V_2^2}{R_L} = 1 \ W$ .

Circuit 1: Circuit 1 is a voltage divisor. The voltage across the load $V_1 = \dfrac {R_L}{R_{in}+R_L} V_{in} = \dfrac 1{11}(1) = \dfrac 1{11} \ V$ and the load power $P_1 = \dfrac {V_1^2}{R_L} = \dfrac 1{121} \ W$ .

$\implies \dfrac {P_2}{P_1} = \dfrac 1{\frac 1{121}} = \boxed{121}$