a light bulb is connected to a source of emf . There is a 6.20 V drop across the

Question

a light bulb is connected to a source of emf . There is a 6.20 V drop across the light bulb, and a current of 4.1 A flowing through the light bulb

a) what is the resistance of the light bulb ?

b) A second light bulb, identical to the first, is connected in series with first bulb, The potential drop across the bulbs is now 6.29 V, and the current through the bulbs is 2.9 A. Calculate the resistance of each light bulb

c) why are your answers to part (a) and (b) not the same?

Answer: a) 1.5

              b) 1.1

Explanation / Answer

We know that V = RI

a) 6.20 = 4.1 * R so R = 1.51

b) When resistors are connected in series, the current through them is the same, and the voltage through them is equal to the sum of the individual voltages. So

Vtotal = IR1 + IR2

Since R1 and R2 are equivilant, we get

Vtotal = IR + IR = 2IR

which becomes

6.29 = 2*2.9*R or R = 1.08

c) In the second case, the bulbs are connected in series, causing the voltage to change. Also, the current is changed. As such, the resistances become different

I hope this helped!

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