Question

when two resistors are wired in series with a 12 v battery, the current through the battery is 0.31 a. when they are wired in parallel with the same battery, the current is 1.60 a.

Answer 1

Let  R₁  and R₂ be the  two Resistance

R₁ - Resistance of 1st resistor

R₂ - Resistance of 2nd resistor

V = Voltage = 12V

I = O.31A ( in series)

I  = 1.6 A ( in parallel)

when two resistance connected in series

Rs = R₁ + R₂

V = IRs = 12 /0.31 V

R₁+R₂ = 38.700Ω   (equation1.)

When two resistance connected in parallel

    $R_{p} =\frac { R_{1} R_{2}}{R_{1} + R_{2} }$

V = I Rp

V=  I (R₂ R₁ /R₁ +R₂)

R₁ R₂ = 290.25 Ω  

As we know

(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2

Using above values we get,

 (R₁ - R₂ ) ²  = (38.70) ²  - 4x 290.25

 (R₁ - R₂ ) ²   = 336.69  Ω  

R₁ - R₂ = 18.34     (equation 2 )

Using equation 1 and 2 we get

R₁+ R₂ = 38.70 52

R₁ - R₂ = 18.34

R₁ = 28.535 Ω  

Using the value of  R₁ in equation 1 we get

R₂ = 38·700 -  28.535

RR₂ = 10.125 Ω  

R₁ = 28. 535 v  and   R₂  10.125 Ω  

To know more about resistance in series and parallel:

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