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Two electric bulb P and Q have their resistance are in 1:2.they are connected in series across a battery. Find the power dissipation in these bulb |
| Answer» HAT, the ratio of the resistance of two electric bulbs P and Q are 1:2. Both of them (bulbs) are connected in a series across a battery. MEANS current is the same in both the bulbs.The resistance of bulb P, resistance is 1Ω and for bulb Q, resistance is 2Ω.Now, P' = VI { P' = V (Q/t) = VI }Here, Power is denoted by P'.Also, from Ohm's LawV = IRSo, P' = I²R{ P' = Power, I = Current and R = Resistance }For bulb P: (Power = P1)→ P1 = I²(1)For bulb Q: (Power = P2)→ P2 = I²(2)So, → P1/P2 = (I² × 1)/(I² × 2)→ P1/P2 = 1/2The power dissipation in these bulbs is the same or equal to the ratio of resistance OFFERED i.e. 1:2. | |