I have the following circuit (below). In order to calculate the current I1 as a function of the resistors and the voltage of the circuit, I'm doing the following:

$$R_{eq} = (R1 //R2 +R3); R_{eq}I = V \iff I = \frac{V}{R_{eq}} = \frac{V}{R3 + \frac{R1+R3}{R1R2}}$$

I know that

$$V = 1 V, R1 = R2 = R3 = 2 \Omega$$

So the current should be

$$I = \frac{1}{2 + 1} = .333$$

However, the textbook gives the expression for I (Agarwal and Lang(2005) Foundations of Analog and Digital, page 152):

$$R_{eq} = (R1 //R2 +R3); R_{eq}I = V \iff I = = \frac{1V}{2\Omega + 2\Omega}$$

I don't get what I did wrong. Those resistors R2 and R1 are in parallel so wouldn't it be the equivalent resistance R1//R2 + R3?


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


If the 1V is the voltage across the I1 then your answer is correct. There's certainly no way that the book's answer is right. Probably just an error in the book, happens all the time.

Some publishers will release 'errata' for textbooks with corrections for errors on their websites.

  • 2
    \$\begingroup\$ Yes, it's an error in the book. Even if they were measuring the voltage elsewhere in the circuit I can see nowhere where that answer applies. \$\endgroup\$ Commented Jul 19, 2022 at 9:12

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