Question

When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same emf is placed across the same two resistors in parallel, the current through the emf is 10 A. What is the magnitude of the greater of the two resistances

Answer 1

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

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

Answer 2

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

$\frac{R_{1}R_{2}}{R_{1}+R_{2}}=2$    .... (2)

From equation (1) and equation (2), we get

R1 x R2 = 20

Put in equation (1), we get

$R_{1}^{2}-10R_{1}+20 = 0$

$R_{1}=\frac{10\pm \sqrt{100-80}}{2}$

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.