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.
Answer:
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:
[tex]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[/tex]
- 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:
[tex]R_{equiv} = R_{1} + R_{2//3} = 20 \Omega + 7.5 \Omega = 27.5 \Omega[/tex]
- 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:
[tex]I =\frac{V}{R_{equiv} } =\frac{22 V}{27.5 \Omega} = 0.8 A[/tex]
- Now, we can find the voltage V₁ through R₁, just by applying Ohm's Law to R₁, as follows:
[tex]V_{1} = I* R_{1} = 0.8 A * 20 \Omega = 16 V[/tex]
- Repeating for the parallel combination of R₂ and R₃, we get:
[tex]V_{2//3} = I* R_{2//3} = 0.8 A * 7.5 \Omega = 6 V[/tex]
- 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:
[tex]I = \frac{V}{R_{1} } = \frac{22 V}{20 \Omega} = 1.1 A[/tex]
- Applying Ohm's law to R₁, we get the voltage through it, as follows:
[tex]V_{1} = I* R_{1} = 1.1 A * 20 \Omega = 22 V[/tex]
- So the final voltages through the 3 resistors will be as follows:
V₁ = 22.0 V
V₂ =0.00 V
V₃ = 0.00 V
