Question

A resistor r1 dissipates the power p when connected

Answer 1

Answer:

A resistor R1 dissipates the power P when connected to a certain generator. If the resistor R2 is put in series with R1, the power dissipated by R1. Since in the second case the resistance (R1+R2) is higher than that in the first case (R1). Therefore power dissipation in the second case will be decreased .

Answer 2

Given that,

First resistor = R₁

Dissipate power = p

Suppose, A resistor r₁ dissipates the power p when connected to a certain generator. If the resistor R₂ is put in series with R₁, the power dissipated by R₁

Let the generator produced  voltage V.

Dissipated power :

The dissipated power is equal to the square of voltage divided by the resistance.

In mathematically,

$P=\dfrac{V^2}{R}$

Where, V = voltage

R = Resistance

So, The dissipated power due to first resistor

$P=\dfrac{V^2}{R_{1}}$

If the the resistor R₂ is put in series with R₁, then

The resistance in the circuit is

$R=R_{1}+R_{2}$

We need to calculate the dissipated power

Using formula of dissipated power

$P=\dfrac{V^2}{R}$

Put the value of R into the formula

$P=\dfrac{V^2}{R_{1}+R_{2}}$

Here, voltage is constant and resistance is increasing

Hence, The power dissipated will be decreases.