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I am confused by a claim in my textbook which states that the two circuits shown below are equivalent:

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My concept on this is not good. My understanding on circuit (c) is like what I write. For circuit (a), does the right side of the resistor to be 20V and left side be 0V? If I am wrong, how do I understand it? Hope someone can clear my concept about this.

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closed as off-topic by John Rennie, Chris, freecharly, Jon Custer, AccidentalFourierTransform Mar 5 '18 at 2:07

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We call those circuits 'equivalent' because their behaviour is identical in terms of the voltage and current they deliver to an external load $R$ connected to the terminals on the right of each circuit, independently of the value of $R$.

So, let's take a look:

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  • For the circuit on the left, if we connect a load $R_L$ to the terminals, then the voltage across that load will be $$V_\mathrm{out} = \frac{R_L}{R_L+R_0}V_0,$$ where $R_0=5\:\Omega$ and $V_0= 20\:\mathrm{V}$.

  • For the circuit on the right, if we connect a load $R_L$ to the terminals, it will draw a current $I_L$ at a voltage $V_\mathrm{out}=I_LR_L$, which must match the voltage $I_0R_0=V_\mathrm{out}=I_LR_L$ over the internal resistance, with the sum of these matching the source current, i.e. $$I_\mathrm{source} = I_0 + I_L = \frac{V_\mathrm{out}}{R_0} + \frac{V_\mathrm{out}}{R_L} = \frac{R_0+R_L}{R_0R_L}V_\mathrm{out},$$ or in other words $$V_\mathrm{out} = \frac{R_LR_0}{R_L+R_0}I_\mathrm{source} = \frac{R_L}{R_L+R_0}R_0I_\mathrm{source},$$ where $R_0I_\mathrm{source} = \rm 4\:A \times 5\:\Omega = 20\:V$.

In other words, no matter what load resistance we connect to the source, it will deliver the same voltage and current, and we won't be able to distinguish which of the two configurations is on the left-hand side of the terminals. This includes the limit of an open circuit (i.e. $R_L\to\infty$, which is what the currently-accepted answer restricts itself to), but it is much broader than that, including finite loads as well as the circuits' behaviour under a short-circuit induced between the terminals, as the $R_L\to0$ limit.

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See, whenever there is a potential difference across a resistor, current will flow through it. Circuit A corresponds to an open circuit. Hence no current will flow through it as there is no potential difference across the resistor! Second thing- In circuits we consider only potential 'difference' and not potential. Hence in circuit C, you can take any two values a and b which give a-b=20V. Considering a different value will not affect your overall calculations. Hope it helps.

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