1. **Problem statement:** We have a circuit with three lamps X, Y, and Z connected as shown. The emf of the cell is 20V with negligible internal resistance. The power dissipated by lamps X, Y, and Z are 10W, 20W, and 20W respectively. We need to find the voltage across Lamp X and Lamp Y. 2. **Relevant formulas:** Power dissipated by a resistor (or lamp) is given by $$P = \frac{V^2}{R}$$ where $P$ is power, $V$ is voltage across the lamp, and $R$ is resistance. 3. **Step 1: Calculate resistance of each lamp using power and voltage:** - Let $V_X$, $V_Y$, and $V_Z$ be voltages across lamps X, Y, and Z respectively. - The total voltage from the cell is 20V. 4. **Step 2: Analyze the circuit:** - Lamp X is in series with the parallel combination of lamps Y and Z. - Voltage across X plus voltage across parallel combination of Y and Z equals 20V: $$V_X + V_{YZ} = 20$$ - Since Y and Z are in parallel, voltage across Y equals voltage across Z: $$V_Y = V_Z = V_{YZ}$$ 5. **Step 3: Express resistances in terms of voltage and power:** - For lamp X: $$R_X = \frac{V_X^2}{P_X} = \frac{V_X^2}{10}$$ - For lamp Y: $$R_Y = \frac{V_Y^2}{20}$$ - For lamp Z: $$R_Z = \frac{V_Z^2}{20} = \frac{V_Y^2}{20}$$ 6. **Step 4: Calculate equivalent resistance of Y and Z in parallel:** $$\frac{1}{R_{YZ}} = \frac{1}{R_Y} + \frac{1}{R_Z} = \frac{1}{\frac{V_Y^2}{20}} + \frac{1}{\frac{V_Y^2}{20}} = 2 \times \frac{20}{V_Y^2} = \frac{40}{V_Y^2}$$ $$R_{YZ} = \frac{V_Y^2}{40}$$ 7. **Step 5: Total resistance in series:** $$R_{total} = R_X + R_{YZ} = \frac{V_X^2}{10} + \frac{V_Y^2}{40}$$ 8. **Step 6: Current through the circuit:** - Current $I$ is same through series elements: $$I = \frac{20}{R_{total}}$$ - Also, current through X is: $$I = \frac{V_X}{R_X} = \frac{V_X}{\frac{V_X^2}{10}} = \frac{10}{V_X}$$ 9. **Step 7: Equate currents:** $$\frac{20}{\frac{V_X^2}{10} + \frac{V_Y^2}{40}} = \frac{10}{V_X}$$ 10. **Step 8: Simplify equation:** Multiply both sides by denominator: $$20 = \frac{10}{V_X} \left( \frac{V_X^2}{10} + \frac{V_Y^2}{40} \right)$$ $$20 = \frac{10}{V_X} \times \frac{V_X^2}{10} + \frac{10}{V_X} \times \frac{V_Y^2}{40}$$ $$20 = V_X + \frac{V_Y^2}{4 V_X}$$ 11. **Step 9: Use voltage sum:** $$V_X + V_Y = 20$$ So, $$V_Y = 20 - V_X$$ 12. **Step 10: Substitute $V_Y$ into equation:** $$20 = V_X + \frac{(20 - V_X)^2}{4 V_X}$$ 13. **Step 11: Multiply both sides by $4 V_X$ to clear denominator:** $$80 V_X = 4 V_X^2 + (20 - V_X)^2$$ 14. **Step 12: Expand and simplify:** $$(20 - V_X)^2 = 400 - 40 V_X + V_X^2$$ So, $$80 V_X = 4 V_X^2 + 400 - 40 V_X + V_X^2$$ $$80 V_X = 5 V_X^2 + 400 - 40 V_X$$ 15. **Step 13: Rearrange terms:** $$5 V_X^2 + 400 - 40 V_X - 80 V_X = 0$$ $$5 V_X^2 - 120 V_X + 400 = 0$$ 16. **Step 14: Divide entire equation by 5:** $$V_X^2 - 24 V_X + 80 = 0$$ 17. **Step 15: Solve quadratic equation:** $$V_X = \frac{24 \pm \sqrt{24^2 - 4 \times 80}}{2} = \frac{24 \pm \sqrt{576 - 320}}{2} = \frac{24 \pm \sqrt{256}}{2} = \frac{24 \pm 16}{2}$$ 18. **Step 16: Calculate roots:** $$V_X = \frac{24 + 16}{2} = 20$$ or $$V_X = \frac{24 - 16}{2} = 4$$ 19. **Step 17: Check for valid voltage:** - If $V_X = 20$, then $V_Y = 0$ which is not possible since lamps Y and Z dissipate power. - So, $V_X = 4$ volts. - Then $V_Y = 20 - 4 = 16$ volts. **Final answer:** - Voltage across Lamp X is **4V**. - Voltage across Lamp Y is **16V**.