A 75 watt light bulb dissipates 75 W of power when connected to an emf of 120 V.

Question

A 75 watt light bulb dissipates 75 W of power when connected to an emf of 120 V. A 100 watt light bulb dissipates 100 W of power when connected to an emf of 120 V.

a) What is the resistance of the 75 W bulb and the 100 W bulb?

b) A 75 W bulb and a 100 W bulb are connected in parallel to a 120 V emf. What is the power dissipated by each bulb? Thus, which bulb is brighter?

c) A 75 W bulb and 100 W bulb are connected in series to a 120 emf. What is the power dissipated by each bulb? Thus, which bulb is brighter?

Explanation / Answer

Data: Power of the first bulb, P1 = 75 W Power of the second bulb, P2 = 100 W Emf, E = 120 V Solution: (a) Resistance of first bulb, R1 = E^2 / P1                                           = 120^2 / 75                                           = 192 Resistance of second bulb, R2 = E^2 / P2                                                = 120^2 / 100                                               = 144 Ans: Resistance of first bulb, R1 = 192 Resistance of second bulb, R2 = 144 (b) Ans: Power dissipated by first bulb, P1 = 75 W Power dissipated by second bulb, P2 = 100 W The second bulb of 100 W glows brighter. (c) Ratio of emf's across first and second bulbs, E1:E2 = R1:R2 So, emf applied to first bulb, E1 = [ R1 / ( R1 + R2) ] * E                                                  = [ 192 / ( 192 + 144 ) ] * 120                                                  = 68.6 V Emf applied to second bulb, E2 = 120 - 68.6                                                  = 51.4 V Power dissipated by first bulb, P1 = E1^2 / R1                                                     = 68.6^2 / 192                                                     = 24.5 W Power dissipated by second bulb, P2 = E2^2 / R2                                                          = 51.4^2 / 144                                                          = 18.35 W Ans: Power dissipated by first blb, P1 = 24.5 W Power dissipated by second bulb, P2 = 18.35 W The first bulb glows brighter. Resistance of second bulb, R2 = E^2 / P2                                                = 120^2 / 100                                               = 144 Ans: Resistance of first bulb, R1 = 192 Resistance of second bulb, R2 = 144 (b) Ans: Power dissipated by first bulb, P1 = 75 W Power dissipated by second bulb, P2 = 100 W The second bulb of 100 W glows brighter. (c) Ratio of emf's across first and second bulbs, E1:E2 = R1:R2 So, emf applied to first bulb, E1 = [ R1 / ( R1 + R2) ] * E                                                  = [ 192 / ( 192 + 144 ) ] * 120                                                  = 68.6 V Emf applied to second bulb, E2 = 120 - 68.6                                                  = 51.4 V Power dissipated by first bulb, P1 = E1^2 / R1                                                     = 68.6^2 / 192                                                     = 24.5 W Power dissipated by second bulb, P2 = E2^2 / R2                                                          = 51.4^2 / 144                                                          = 18.35 W Ans: Power dissipated by first blb, P1 = 24.5 W Power dissipated by second bulb, P2 = 18.35 W The first bulb glows brighter.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.