Using ohm's law and the characteristics of series and parallel circuits, we find that the correct answer is B: Voltage on the voltmeter 1.5 V
given parameters
* The element X has a resistance Rₓ = R
* Element Y with resistance R_y = R / 2
* The voltmeter with internal resistance R_v = R
* Battery voltage V = 6V
* the circuit is in the diagram
To find
The Voltmeter reading.
For this exercise we use ohm's law which establishes a linear relationship between voltage and current
V = i R
Let's solve the circuit in parts:
1 part. We reduce the part in parallel, finding its equivalent resistance
R_ {eq1}
The equivalent resistance between Y element and the voltmeter that is in parallel, therefore their equivalent resistance is
[tex]\frac{1}{R_{eq1} } = \frac{1}{R_y} + \frac{1}{R_v}[/tex]
[tex]\frac{1}{R_{eq1} }= \frac{1}{\frac{R}{2} } + \frac{1}{R} = \frac{3}{R}[/tex]
[tex]R_{eq1} = \frac{R}{3}[/tex]
Now element X and this equivalent resistance are in series ( see attached), so the equivalent resistance of the entire circuit is
R_{eq} = Rₓ + R_{eq1}
we substitute
R_{eq} = [tex]R + \frac{R}{3} = \frac{4}{3} \ R[/tex]
2 part. We look for the current of the circuit, usin the ohm's Law
V = i R_{eq}
i = [tex]\frac{V}{R_{eq} }[/tex]
i = [tex]\frac{6}{\frac{4R}{3} } = \frac{18}{4R}[/tex]
i = 4.5 / R
In a series circuit the current is constant and the total voltage is the sum voltage of each element, so the voltage in the parallel R_{eq1}
V₂ = I R_{eq1}
V₂ = [tex]\frac{4.5}{R} \ \frac{R}{3}[/tex]4.5 / R R / 3
V₂ = 1.5 V
3 part. The voltmeter and the Y element are in parallel, so the voltage in the two elects is the same, consequently the voltage of the voltmeter (V_v)
V_v = V₂ = 1.5 V
Using ohm's law and the properties of series and parellel circuits we find the voltage across the voltmeter
V_v = 1.5 V
To learn about series and parallel circuits here: brainly.com/question/11409042
