What would the voltages across each of the resistors in the preceding circuit be if a wire were placed across the ends of resistor 3 (that is, resistor 3 is shorted out)

Complete Question:

What would the voltages across each of the resistors in the preceding circuit be if a wire were placed across the ends of resistor 3 (that is, resistor 3 is shorted out)

R1 = 20 Ω, R2 = 30 Ω, and R3 = 10 Ω.

V₁ = 22.0 V  V₂=V₃=0.00 V

Explanation:

Prior to apply Ohm's law to the series circuit attached, we need to find the equivalent resistance for the parallel combination of R₂ and R₃, as follows:

     

Now, in a series circuit, the equivalent resistance is equal to the sum of all the resistances in series, so in this case we get:

           

Now, we can apply Ohm's Law to the circuit in order to find the current I (which is the same at any point in the circuit due to the charge conservation principle), as follows:

       

Now, we can find the voltage V₁ through R₁, just by applying Ohm's Law to R₁, as follows:

       

Repeating for the parallel combination of R₂ and R₃, we get:

       

Now, if we short out R₃ (and consequently R₂ too as both are in parallel each other), we would build a new combination of resistances in parallel, which value will be 0 Ω, due to one of them (the wire) has zero resistance.So, we will have V₂ = V₃ = 0 V.So, the only resistor in the circuit will be just R₁ = 20Ω.The new value of the current will be simply the battery voltage (fully applied through R₁) divided by R₁, as follows:

       

Applying Ohm's law to R₁, we get the voltage through it, as follows:

       

So the final voltages through the 3 resistors will be as follows:

       V₁ = 22.0 V

       V₂ =0.00 V

       V₃ = 0.00 V

Rain is pulled by gravity which is a non contact force