Two resistors R1 and R2 of 4Ω and 6Ω are connected in parallel across a battery. The ratio of power dissipated in them, P1: P2 will be.

To find the ratio of power dissipated in the resistors R1 and R2, we can use the formula for power dissipation in resistors connected in parallel. The power dissipated in a resistor is given by:

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

where $P$ is the power, $V$ is the voltage across the resistor, and $R$ is the resistance.

Since the resistors are connected in parallel, the voltage across each resistor is the same. Let $V$ be the voltage across each resistor.

For resistor R1 (4Ω): $P1 = \frac{V^2}{R1} = \frac{V^2}{4}$

For resistor R2 (6Ω): $P2 = \frac{V^2}{R2} = \frac{V^2}{6}$

To find the ratio of power dissipated in R1 to R2, we take the ratio of P1 to P2: $\frac{P1}{P2} = \frac{\frac{V^2}{4}}{\frac{V^2}{6}} = \frac{V^2}{4} \times \frac{6}{V^2} = \frac{6}{4} = \frac{3}{2}$

Therefore, the ratio of power dissipated in them, $P1 : P2$, is: $3 : 2$