The question asks; "P14.25 Analyze each of the ideal-op-amp circuits shown in Figure P14.25 to find expressions for 𝑖𝑜. The top part is the question, the bottom is the solution from the solutions manual. I don't understand how they apply KVL between the inputs since there is no wire connecting them, are we allowed to apply KVL by jumping between wires?
They write 𝑉1 = 0 + 𝑅𝑖0 + 0 + 𝑉2, and write 0V at each input, adding the voltages between the inputs.
Negative feedback causes the op-amp to adjust its output potential to whatever is necessary to equalise its two input potentials. That's the principle behind the "virtual ground" you read about. Here's an answer I wrote explaining this behaviour.
Each op-amp in your circuit has negative feedback, and therefore each opamp's inverting and non-inverting inputs have equal potential. If I redraw the circuit with some node names, it's easier to show:
simulate this circuit – Schematic created using CircuitLab
Negative feedback produce these conditions:
$$ V_B = V_A = V_1 $$ $$ V_C = V_D = V_2 $$
The zero terms in the KVL equation you wrote correspond to the potential differences \$V_A-V_B\$ and \$V_D-V_C\$, which are both zero due to op-amp behaviour when there's negative feedback.
Therefore, the voltage across resistor R2 is:
$$ V_{R2} = V_B - V_C = V_1 - V_2 $$
It's not KVL that makes the voltages across each pair of differential inputs zero, it's an effect of negative feedback and extremely high gain.
It's also an approximation only and is invalid if the op-amps saturate (with voltage or they can't supply the current required) or if the feedback is positive rather than negative.
If the op-amp gain is, say, 200,000 then the DC voltage change between the inputs to drive the output would be 1/200,000 of the required output voltage change. If the supplies are (say) +/-5V that's +/-25uV. In practice there's also an offset voltage which might be some mV or uV depending on the op-amp design.
simulate this circuit – Schematic created using CircuitLab
The circuit presented by OP is a constant current source that supplies a "floating" load and a current-setting resistor R in series. The voltage across the resistor is formed as a difference between two voltages VB and VD, which are "copies" of the input voltages VA and VC (see the schematics below). This creates the illusion that the input voltages are as if connected to the resistor.
This circuit phenomenon is known by the figurative name "virtual short" which suggests what the idea is here. This is a concept that can be demonstrated with just a few voltage sources, without any op-amps. Let's see how.
If we connect a voltage source VA (point A) to the left end of a right-grounded resistor R (point B), there is a "real short" (wire) between them, and the voltage at point B follows the voltage at point A. So VB and accordingly IB are constant.
But we can achieve the same result without the two sources actually being connected. For this purpose, we can connect (unnoticeable to others :-) another voltage source VB to point B and adjust its voltage equal to the voltage of VA. This creates the illusion that the source VA is somehow connected (via the dashed line in violet) to point B and it determines its voltage. So VA, VB and accordingly IR = VB/R are constant.
In the same "virtual way", we can "connect" the right end (point D) of the left-grounded resistor R to the input source VC (point C). As above, we connect another voltage source VD to point D and adjust its voltage equal to the voltage of VC. This creates the illusion that the source VC is connected to point D and it determines its voltage. So VC, VD and accordingly IR = VC/R are constant.
In this way, the resistor R is virtually connected (via the two dashed lines in violet) between the two input voltage sources VA and VC (V1 and V2 in the OP's schematic). Thus the voltage across it is VR = VA - VC, and the current through it is I = (VA - VC)/R.
The load connected in series to the resistor R does not change anything. The op-amp overcomes it and keeps VB = VA.
Finally, let's examine the operation of the OP's circuit. For concreteness, assume that V1 > V2.
RL = 0: Let's initially the load resistance RL = 0. Then, the circuit consists of two op-amp followers whose outputs are connected through a resistor R. The first op-amp (OA1) keeps the voltage at the left end (its inverting input) of the resistor equal to V1; the second op-amp (OA2) keeps the voltage at the right end (its inverting input) equal to V2. So, the current through the load and R is I = (V1 - V2)/R; it exits OA1's output and enters OA2's input.
RL increases: The current tries to decrease since the total resistance R + RL increases while, at the first moment, the total voltage VOA1 - VOA2 does not change. OA1 senses the voltage drop V- at its inverting input and begins increasing its output voltage until it restores V-. Thus the circuit behaves as a non-inverting amplifier with changing resistor R2. As a result, the voltage drop VR and accordingly the current does not change.
See also my story about voltage follower.
are we allowed to apply KVL by jumping between wires?
Yes of course. At every element along the KVL path, we jump from the wire connected to one terminal to the wire connected the other terminal. The op-amp is no different. Always be wary of the resistance element \$R_{in}\$ between the input terminals of the op-amp
simulate this circuit – Schematic created using CircuitLab
Figure 1 demonstrates the input KVL path including the op-amp input resistances. Across these input elements is a voltage \$v_d=\frac{V_{out}}{A_VOL}\$ that is the difference between the input terminal voltages of each 0p-amp. Often \$v_d\$ is considered 0V if \$A_{VOL}\$ is large enough that its value is insignificant.
So just jumping between wires is the wrong visualization. There is always a passive element there even if its influence is insignificant.
So for a clockwise KVL path starting from the ground node: $$ V_1-v_{d1}-V_R+v_{d2}-V_2=0 $$
So if \$A_{VOL}\$ is large enough so that \$v_{d1}\approx v_{d2}\approx 0\$ resulting in:
$$ V_R=V_1-V_2 $$
