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b) State the laws of reflection of light.

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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 2Ω2 \Omega are first connected in series and then in parallel in an electric circuit to a potential difference of 6 V6 \text{ V}. Find the electric power of this circuit in series and parallel combinations.

Given:

  • Resistance of each resistor, R1=2ΩR_1 = 2 \Omega
  • Resistance of each resistor, R2=2ΩR_2 = 2 \Omega
  • Potential difference, V=6 VV = 6 \text{ V}

Case 1: Resistors connected in series

  1. Calculate the equivalent resistance (RsR_s) for resistors in series.

    Rs=R1+R2R_s = R_1 + R_2 Rs=2Ω+2ΩR_s = 2 \Omega + 2 \Omega Rs=4ΩR_s = 4 \Omega

  2. Calculate the electric power (PsP_s) using the formula P=V2RP = \frac{V^2}{R}.

    Ps=V2RsP_s = \frac{V^2}{R_s} Ps=(6 V)24ΩP_s = \frac{(6 \text{ V})^2}{4 \Omega} Ps=36 V24ΩP_s = \frac{36 \text{ V}^2}{4 \Omega} Ps=9 WP_s = 9 \text{ W}

    The electric power in the series combination is 9 W\boxed{\text{9 W}}.

    Case 2: Resistors connected in parallel

  3. Calculate the equivalent resistance (RpR_p) for resistors in parallel.

    1Rp=1R1+1R2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} 1Rp=12Ω+12Ω\frac{1}{R_p} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} 1Rp=22Ω\frac{1}{R_p} = \frac{2}{2 \Omega} 1Rp=11Ω\frac{1}{R_p} = \frac{1}{1 \Omega} Rp=1ΩR_p = 1 \Omega

  4. Calculate the electric power (PpP_p) using the formula P=V2RP = \frac{V^2}{R}.

    Pp=V2RpP_p = \frac{V^2}{R_p} Pp=(6 V)21ΩP_p = \frac{(6 \text{ V})^2}{1 \Omega} Pp=36 V21ΩP_p = \frac{36 \text{ V}^2}{1 \Omega} Pp=36 WP_p = 36 \text{ W}

    The electric power in the parallel combination is 36 W\boxed{\text{36 W}}.

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