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Home/ Questions/An ammeter is connected in series to a battery of voltage Vb and a resistor of...
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lyytutoria
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lyytutoriaExpert
Asked: April 10, 20222022-04-10T07:48:16+00:00 2022-04-10T07:48:16+00:00In: Physics

An ammeter is connected in series to a battery of voltage Vb and a resistor of…

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An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru (Figure 1) . The ammeter reads a current Io. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter’s reading drops to I1. Finally, a second resistor, also of resistance Rr, is connected in series as well. Now the ammeter reads I2.

If I1/I0=4/5, find I2/I0.

Express the ratio I2/I0 numerically.

I2/I0 =

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An ammeter is connected in series to a battery of voltage Vb and a resistor of...


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Light is made up of transverse electrical electromagnetic field oscillations. When we speak of light oscillations, it’s just about electrical field oscillations. The process of polarization involves a way of limiting light oscillations to one plane. This plane is called the vibration plane. The other plane parallel to this plane that contains the direction of the propagation is known as the polarization plane in which the electric field is blocked from oscillation or from oscillating. Intensity of plane polarized light passing through a polarizer is given by the equation I=I0cos2(θ). Here I0, I, θ are the incident intensity, transmitted intensity, angle between the plane of vibration of the polarized beam and transmission axis of the polarizer respectively.

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    lyytutoria Expert
    2022-04-12T16:02:15+00:00Added an answer on April 12, 2022 at 4:02 pm

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  2. tonytutoria
    2022-04-10T07:48:18+00:00Added an answer on April 10, 2022 at 7:48 am

    Find I2 /I0 if I1 /I0 = 4/5. The ratio I2/I0 can be expressed numerically.

    We can also say that I1 /I0 = 4/5. I1 = 4, I0 = 5. We also know V = IR, and V/R. Ru is 1. This means that I0 = V / 1 equals 5 so V = 5. For I1:

    V = IR

    So the resistors are in series.

    5 = 4 * (Ru + Rr)

    5 = 4 * (1 + Rr)

    Rr = 0.25

    This can be applied to I2 (again the resistors in series are added so we add them).

    V = I2 *R

    V = I2 *(Rr + Rr+ Ru)

    5 = I2 (1.5)

    I2 = 3.333

    So I2 = I0 = 3.333/5 = 0.667

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