b) State the laws of reflection of light.

Here are the solutions for question 12: b) State the laws of reflection of light. The laws of reflection of light are: 1. The incident ray, the reflected ray, and the normal to the surface at the point of incidence all lie in the same plane. 2. The angle of incidence is equal to the angle of reflection. 12. Two resistors having resistance $2 \Omega$ are first connected in series and then in parallel in an electric circuit to a potential difference of $6 \text{ V}$. Find the electric power of this circuit in series and parallel combinations. Given: Resistance of each resistor, $R_1 = 2 \Omega$ Resistance of each resistor, $R_2 = 2 \Omega$ Potential difference, $V = 6 \text{ V}$ Case 1: Resistors connected in series Step 1: Calculate the equivalent resistance ($R_s$) for resistors in series. $$R_s = R_1 + R_2$$ $$R_s = 2 \Omega + 2 \Omega$$ $$R_s = 4 \Omega$$ Step 2: Calculate the electric power ($P_s$) using the formula $P = \frac{V^2}{R}$. $$P_s = \frac{V^2}{R_s}$$ $$P_s = \frac{(6 \text{ V})^2}{4 \Omega}$$ $$P_s = \frac{36 \text{ V}^2}{4 \Omega}$$ $$P_s = 9 \text{ W}$$ The electric power in the series combination is $\boxed{\text{9 W}}$. Case 2: Resistors connected in parallel Step 1: Calculate the equivalent resistance ($R_p$) for resistors in parallel. $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\frac{1}{R_p} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega}$$ $$\frac{1}{R_p} = \frac{2}{2 \Omega}$$ $$\frac{1}{R_p} = \frac{1}{1 \Omega}$$ $$R_p = 1 \Omega$$ Step 2: Calculate the electric power ($P_p$) using the formula $P = \frac{V^2}{R}$. $$P_p = \frac{V^2}{R_p}$$ $$P_p = \frac{(6 \text{ V})^2}{1 \Omega}$$ $$P_p = \frac{36 \text{ V}^2}{1 \Omega}$$ $$P_p = 36 \text{ W}$$ The electric power in the parallel combination is $\boxed{\text{36 W}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.