Two resistors

Two resistors when they give 25 ohms in series and 4 ohms in parallel what the values of

Correct result:

R1 =  20 Ω
R2 =  5 Ω

Solution:

a+b=25 ab/(a+b)=4 a(25a)/25=4  0.04a2+a4=0 0.04a2a+4=0  p=0.04;q=1;r=4 D=q24pr=1240.044=0.36 D>0  a1,2=q±D2p=1±0.360.08 a1,2=12.5±7.5 a1=20 a2=5   Factored form of the equation:  0.04(a20)(a5)=0  R1=a1=20=20 Ωa+b=25 \ \\ ab/(a+b)=4 \ \\ a(25-a)/25=4 \ \\ \ \\ -0.04a^2 +a -4=0 \ \\ 0.04a^2 -a +4=0 \ \\ \ \\ p=0.04; q=-1; r=4 \ \\ D=q^2 - 4pr=1^2 - 4\cdot 0.04 \cdot 4=0.36 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 1 \pm \sqrt{ 0.36 } }{ 0.08 } \ \\ a_{1,2}=12.5 \pm 7.5 \ \\ a_{1}=20 \ \\ a_{2}=5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 0.04 (a -20) (a -5)=0 \ \\ \ \\ R_{1}=a_{1}=20=20 \ Ω

Checkout calculation with our calculator of quadratic equations.

R2=a2=5=5 ΩR_{2}=a_{2}=5=5 \ Ω



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