two resistors R1=-4 ohms and R2=5 ohms, are connected parallel across a potential difference. if P1 and P2 represents the power dissipate in R1 and R2 respectively, then the ratio P1:P2 is?

P = E^2/R

P1:P2 = E^2/4 : E^2/5 = 5:4

To find the ratio of power dissipated in resistors R1 and R2, we can use the formula for power dissipation in a resistor:

P = (V^2) / R

Where P is power, V is potential difference, and R is resistance.

Given that R1 = -4 ohms and R2 = 5 ohms, we know that resistances cannot be negative. Therefore, we assume that R1 = 4 ohms.

Since the resistors are connected in parallel, they have the same potential difference across them.

Let's assume the potential difference is V.

For resistor R1:
P1 = (V^2) / R1 = V^2 / 4

For resistor R2:
P2 = (V^2) / R2 = V^2 / 5

To find the ratio P1:P2, we divide P1 by P2:

P1 / P2 = (V^2 / 4) / (V^2 / 5)
= (V^2 / 4) * (5 / V^2)
= 5 / 4

Therefore, the ratio P1:P2 is 5:4.

To calculate the power dissipated in each resistor, we can use the formula for power:

P = (V^2 / R),

where P is the power, V is the potential difference applied across the resistor, and R is the resistance.

In this case, we are given two resistors in parallel. When resistors are connected in parallel, the potential difference across each resistor is the same.

So, let's assume the common potential difference across the resistors is V.

For resistor R1 with resistance -4 ohms (assuming it is a typo and should be positive 4 ohms), the power dissipated is given by:

P1 = (V^2 / R1) = (V^2 / 4).

For resistor R2 with resistance 5 ohms, the power dissipated is given by:

P2 = (V^2 / R2) = (V^2 / 5).

To find the ratio P1:P2, we can divide these two equations:

P1/P2 = [(V^2 / 4) / (V^2 / 5)] = 5/4.

Therefore, the ratio P1:P2 is 5:4.