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 pa

Question

1 year ago

13

rallel with the same battery, the current is 1.60 a.

1 answer:

liberstina [14] · Answer 1 · 2024-03-12T00:59:03+03:00

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

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