Question

Two light bulbs have resistances of 40 2 and 80 22. Part B The two light bulbs are connected in series across a 12 V supply. Find the power dissipated in each bulb. Express your answer in watts separated by a comma. ΙΕΙ ΑΣΦ ? P40, P80 = Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Part C The two light bulbs are connected in series across a 12 V supply. Find the total power dissipated in both bulbs. Express your answer in watts. EVO AEO ? P= w Submit Request Answer

Answer 1

The resistances of the light bulbs are

$\\R_{40} = 40 \ \Omega \\R_{80} = 80 \ \Omega$

Part B,

Given, supply is

12 V

When the bulbs are in series, their total resistance is

$R_t = 40 + 80 = 120 \ \Omega$

therefore, current in the circuit is

12 = 0.1 A Rt 120

then, power dissipated in each bulb is

$\\P_{40} = I^2 R_{40} = 0.1^2 * 40 = 0.4 \ W \\\\P_{80} = I^2 R_{80} = 0.1^2 * 80 = 0.8 \ W$

Part C,

The total power dissipated in both bulbs is same as the power output of the supply

therefore,