Equivalent resistance in a parallel circuit

[Kurokari]

Homework Statement

Prove that the equivalent resistance in a parallel circuit with 2 resistors with resistance R1 and R2 is always lesser than the resistance of either resistor.

Homework Equations

R_t= (R1 x R2)/(R1 + R2)

The Attempt at a Solution

I tried starting from R1 + R2 > R1 , where R2 > 0

 

[gneill]

If R_t is smaller than either R1 or R2, then what does that say about the ratios R_t/R1 and R_t/R2?

 

[ehild]

Homework Statement

Prove that the equivalent resistance in a parallel circuit with 2 resistors with resistance R1 and R2 is always lesser than the resistance of either resistor.

Homework Equations

R_t= (R1 x R2)/(R1 + R2)

The Attempt at a Solution

I tried starting from R1 + R2 > R1 , where R2 > 0

That is a good start. The resistances are all positive, R1, R2 and also R_t.

You know that the reciprocal of the parallel resistances add up:

1/R_t=1/R1+1/R2

Multiply the equation by R_t. ...

ehild

 

[willem2]

You can simply compute the difference between R_t and R_1 or R_2 which is easily seen to be positive

\frac {R_1 R_2} {R_1 + R_2} - R_1 =

 

[Kurokari]

EDIT: I've come up to R1/(R1 + R2) > R1 , where R2 > 0

since R1+R2>R1 , then when R1 is divided by R1+R2(a larger number), surely R1/(R1+R2) is lesser than the original number.

now I can't have any reasoning that when I multiply R1/(R1+R2) by R2, tht this inequality still stands.

EDIT:

You can simply compute the difference between R_t and R_1 or R_2 which is easily seen to be positive

\frac {R_1 R_2} {R_1 + R_2} - R_1 =

I see thanks, but isn't it the other way around?

 

[Curious3141]

You can simply compute the difference between R_t and R_1 or R_2 which is easily seen to be positive

\frac {R_1 R_2} {R_1 + R_2} - R_1 =

You must have meant negative, the way that difference was written.

For the thread starter: this is a simple way to do it. Just express the second term over the same denominator, and combine terms and cancel stuff out. The end result should be obviously negative. You can then repeat it for $R_2$.

 

[ehild]

That is a good start. The resistances are all positive, R1, R2 and also R_t.

You know that the reciprocal of the parallel resistances add up:

1/R_t=1/R1+1/R2

Multiply the equation by R_t. ...

ehild

1=R_t/R1+R_t/R2

As both terms on the right-hand side must be positive, both are smaller than 1.

ehild