When two resistors are wired in series with a 12 V battery, the current through the battery is 0.35 A. When they are wired in parallel with the same battery, the current is 1.60 A. What are the values of the two resistors?

The resistances of the two resistors are 23.2 Ω and 11.1 Ω.

Explanation:

Let the resistance of the two resistors be R₁ and R₂

From Ohm's law,

V = IR

where V = voltage on battery = 12 V

I = current through battery

R = equivalent resistance of the resistors in the circuit

When the resistors are connected in series,

I = 0.35 A

R = R₁ + R₂

V = 12 V

12 = 0.35 R

R = (12/0.35)

R = 34.3 Ω

R₁ + R₂ = 34.3 (eqn 1)

When the resistors are connected in parallel,

I = 1.60 A

R = (R₁R₂)/(R₁ + R₂)

V = 12 V

12 = 1.6R

R = 7.5 Ω

(R₁R₂)/(R₁ + R₂) = 7.5

From eqn 1, (R₁ + R₂ = 34.3) and R₂ = 34.3 - R₁

So, we have,

[R₁(34.3 - R₁)]/34.3 = 7.5

34.3R₁ - R₁² = 257.25

R₁² - 34.3R₁ + 257.25 = 0

Solving the quadratic equation,

R₁ = 23.2 Ω or R₁ = 11.1 Ω

If R₁ = 23.2 Ω

R₂ = 34.3 - R₁ = 34.3 - 23.2 = 11.1 Ω

If R₁ = 11.1 Ω

R₂ = 34.3 - R₁ = 34.3 - 11.1 = 23.2 Ω

Hope this helps!!!

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.