Answer:
7.24 ohm
Explanation:
Let R1 and R2 are resistance of two resistors.
Emf=E=20 V
Current,I=2 A
Current,I'=10 A
We have to find the magnitude of the greater of the two resistances.
In series
[tex]R=R_1+R_2[/tex]
[tex]V=IR[/tex]
By using the formula
[tex]20=2(R_1+R_2)[/tex]
[tex]R_1+R_2=\frac{20}{2}=10[/tex]...(1)
In parallel
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}[/tex]
[tex]R=\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]20=10(\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]2=\frac{R_1R_2}{10}[/tex]
[tex]R_1R_2=20[/tex]
[tex]R_2=\frac{20}{R_1}[/tex]
Substitute the value
[tex]\frac{20}{R_1}+R_1=10[/tex]
[tex]R^2_1+20=10R_1[/tex]
[tex]R^2_1-10R_1+20=0[/tex]
[tex]R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}[/tex]
By using quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]R_1=\frac{10\pm 2\sqrt 5}{2}[/tex]
[tex]R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm[/tex]
[tex]R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm[/tex]
Substitute the value
[tex]R_2=\frac{20}{7.24}=2.76 ohm[/tex]
[tex]R_2=\frac{20}{2.76}=7.24 ohm[/tex]
Hence, the magnitude of the greater of the two resistance=7.24 ohm
Answer:
Explanation:
Let the resistance is R1 and R2.
Case 1: When they are connected in series.
Let the equivalent resistance is R.
V = 20 V
i = 2 A
R = V / i = 20 / 2 = 10 ohm
R1 + R2 = 10 ohm ...... (1)
Case 2: When they are connected in parallel.
V = 20 V
i = 10 A
R = 20/10 = 2 ohm
[tex]\frac{R_{1}R_{2}}{R_{1}+R_{2}}=2[/tex] .... (2)
From equation (1) and equation (2), we get
R1 x R2 = 20
Put in equation (1), we get
[tex]R_{1}^{2}-10R_{1}+20 = 0[/tex]
[tex]R_{1}=\frac{10\pm \sqrt{100-80}}{2}[/tex]
Take positive sign
R1 = 7.24 ohm
So, R2 = 10 - 7.24 = 2.76 ohm
So, the greater value of resistance is 7.24 ohm.
