Physics, 22.10.2020 19:01 iamhayls

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)

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Answer from: yfnal3x

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:

$R_{//2-3} = \frac{R_{2} *R_{3}}{R_{2+R_{3} } } = \frac{30 \Omega * 10 \Omega}{30 \Omega + 10 \Omega} = \frac{300 }{40} \Omega = 7.5 \Omega$

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:

$R_{equiv} = R_{1} + R_{2//3} = 20 \Omega + 7.5 \Omega = 27.5 \Omega$

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:

$I =\frac{V}{R_{equiv} } =\frac{22 V}{27.5 \Omega} = 0.8 A$

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

$V_{1} = I* R_{1} = 0.8 A * 20 \Omega = 16 V$

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

$V_{2//3} = I* R_{2//3} = 0.8 A * 7.5 \Omega = 6 V$

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:

$I = \frac{V}{R_{1} } = \frac{22 V}{20 \Omega} = 1.1 A$

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

$V_{1} = I* R_{1} = 1.1 A * 20 \Omega = 22 V$

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

V₁ = 22.0 V

V₂ =0.00 V

V₃ = 0.00 V

What would the voltages across each of the resistors in the preceding circuit be if a wire were plac

Answer from: Quest

whichever one it is, it's definitely not one that you included on the list you provided.

Answer from: Quest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

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