Req of this series/parallel circuit?

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The discussion focuses on calculating the equivalent resistance (Req) of a circuit involving resistors R1, R2, R3, and R4. The initial calculations incorrectly treat R1 and R3 as series, leading to an incorrect Req of 266.7 ohms. A key point raised is the presence of a wire connecting R3, which effectively reduces its resistance to zero when considered in parallel with R3. This results in R3 having no voltage drop, simplifying the circuit significantly. Participants encourage re-evaluating the calculations with the understanding that R3 in parallel with a zero-ohm resistor leads to a new approach for finding the correct Req.
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Homework Statement


What is the Req of this circuit?
The answer should be 150 ohms, but I can't get that.

Homework Equations


Req = R1 + R2
1/Req = 1/R1 + 1/R2

The Attempt at a Solution


If I treat R1 and R3 as series, I get R13= 200 ohms.
Then if I treat R13 and R2 and parallel I get R123 = 66.7 ohms.
Then if I treat R123 and R4 as series, I get Req = 266.7 ohms.

Also, since the current is 0 (open switch), then the potential difference between points P and Q would also be 0, right?
 
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A hint..

There appears to be a wire (grey) connected from one end of R3 to the other end. What does that mean for the voltage across R3 or R3 in general?
 
CWatters said:
A hint..

There appears to be a wire (grey) connected from one end of R3 to the other end. What does that mean for the voltage across R3 or R3 in general?
No clue. It's the same as the source's voltage?

Also, did you manage to get Req?
 
The grey wire has a very low resistance, let's assume it's zero Ohms. You therefore have R3 in parallel with a zero Ohm resistor..

For that bit of the circuit..

I/R = 1/R3 + 1/zero

but 1/zero = ∞

so R3 in parallel with zero Ohms = zero Ohms.

The circuit reduces to..
 

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I'll let you have another go at calculating Req.
 
CWatters said:
The grey wire has a very low resistance, let's assume it's zero Ohms. You therefore have R3 in parallel with a zero Ohm resistor..

For that bit of the circuit..

I/R = 1/R3 + 1/zero

but 1/zero = ∞

so R3 in parallel with zero Ohms = zero Ohms.

The circuit reduces to..
Smart, thanks.
 
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