Resistors in parallel?

Question

From what I understand, the proof is suggesting that $$i(t) = (\frac{1}{R_1}+\frac{1}{R_2})v(t)$$ implies that replacing the two resistors with 1 resistor of value \$\frac{1}{R_1}+\frac{1}{R_2}\$ won't change anything and voltage across the equivalent resistor will be the same.

I don't understand why this is true. $$i(t) = (\frac{1}{R_1}+\frac{1}{R_2})v(t)$$ implies that if you replace the two resistors with 1 resistor of value \$\frac{1}{R_1}+\frac{1}{R_2}\$ and keep the current across the equivalent resistor the same as the sum of the currents flowing across the two initial resistors, the voltage across the new resistor will be equivalent to the voltages across the initial resistors.

In other words, the proof implies if you replace the two resistors with an "equivalent resistor", and force the current flowing through the equivalent resistor \$i(t)\$ then the voltage across the equivalent resistor is \$v(t)\$.

Now in the example above, because of the independent current source, the current across the equivalent resistor is forced to be \$i(t)\$, therefore the voltage across it will also be \$v(t)\$. So for the example above, the proof is valid.

Now consider a more complicated circuit, where you have two resistors in parallel, but no independent current source (but instead its part of a more complicated circuit). Therefore if you replace them with an equivalent resistor, there is no guarantee that the current across the equivalent resistor will be equivalent to the sum of the currents across the original resistors, so the voltage might not necessarily be the same.

So why is this a valid proof for all resistors in parallel instead of only resistors in parallel with a independent current source regulating the current across them?

EDIT: I'm essentially having trouble understanding why the implication of the proof isn't the voltage across the equivalent resistance will be vp only if the current across it is identical to the sum of the currents through the two parallel resistors.

Answer 1

So why is this a valid proof for all resistors in parallel

First, you have an error in your question - the equivalent resistance is

$$R_P = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$

Now, the voltage across the two parallel resistors is what it is regardless of how the voltage comes to be.

However we choose to label that voltage is immaterial, thus, we can arbitrarily label the voltage across the parallel resistors as, e.g., \$v_P\$.

Now, and again, it does not matter how this voltage comes to be, the voltage variable \$v_P\$ is the voltage measured across the parallel resistors when "red" lead is placed on the "\$+\$" labelled terminal and the "black" lead is on the "\$-\$" labelled terminal.

Thus, by Ohm's law, the current through each resistor is

$$i_{R_1} = \frac{v_P}{R_1} $$

$$i_{R_2} = \frac{v_P}{R_2} $$

So, the total current is, by KCL,

$$i_P = i_{R_1} + i_{R_2}$$

and the equivalent resistance is defined as

$$R_P = \frac{v_P}{i_P}$$

thus,

$$R_P = \frac{v_P}{i_{R_1} + i_{R_2}} = \frac{v_P}{\frac{v_P}{R_1} + \frac{v_P}{R_2}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$

Again, if we replace the two parallel resistors with a resistor of resistance \$R_P\$, the current through the equivalent resistance will be identical to the sum of the currents through the two parallel resistors.