The battery has an emf of 3.0 Volts and the light bulbs A and C say 1.5W and B a

Question

The battery has an emf of 3.0 Volts and the light bulbs A and C say 1.5W and B and D
say 4.5W. All light bulbs are made for 3.0V batteries. When the switch is closed:
a. What current was the manufacturer assuming?
b. What are the resistances of bulbs A – D?
c. Calculate the current and Voltage for each of the light bulbs in the circuits.
d. Rank light bulbs A-D by the brightness.
e. Show in the diagram how you would measure the current through light bulb A and
how you would measure the voltage of light bulb D using a multimeter?

(For part a) and b) each different bulb is under manufacturing conditions: ONE bulb connected in series with a 3V battery, so that the bulb has a voltage of 3V across it. The power dissipated by a bulb under manufacturing conditions is NOT the same as the power dissipated by the bulb in the circuit unless the bulb has a potential difference across it equal to the manufacturing voltage.)

For parts c) and d), you need to use the resistances of the bulbs calculated in b)

Explanation / Answer

Given that,

Battery emf = 3 volts

Power rating of the bulbs A and C is P(A) = P(C) = 1.5W and that of B and D is P(B) = P(D) = 4.5 W

(a)Let the current be I.

We know that, P = I2R = V2/R => R =  V2/P

R(A) = R(C) =   (3)2/1.5 = 6 Ohms

R(B) = R(D) = (3)2/4.5 = 2 Ohms

So the current that the manufacture is assuming will be;

I (A) = I(C)= V/R = 3/6 = 0.5 A for the bulbs A and C

I(B)= I(D) = V/R = 3/2 = 1.5 A for the bulbs B and D

part(b)

We know that, P = I2R = V2/R => R =  V2/P

R(A) = R(C) =   (3)2/1.5 = 6 Ohms

R(B) = R(D) = (3)2/4.5 = 2 Ohms

Hence the resistance of bulb A and C is 6 Ohm and that of B and D is 2 Ohm.

(c)Consider the first circuit. In This the bulbs A and B are in series with the battery, Let R1 be the equivalent resistance of the bulbs A and B, so

R1 = R(A) + R(B) = 6 + 2 = 8 Ohm

We know that, V = I R => I = V/R = 3/8 = 0.375 A

We now have the current flowing in the circuit. And we know that, the same current flows through all the resistors connecetd in seris. So

I(A) = I(B) = 0.375 A

V(A) = I(A)xR(A) = 0.375 x 6 = 2.25 Volts

V(B) = I(B)xR(B) = 0.375 x 2 = 0.75 Volts

For the second circuit, we find that the bulbs are connected in parallel, Ler R2 be the equivalent resistance of these bulbs, So

R2 = R(C) x R(D) / [R(C) + R(D) ] = 2 x 6 / (2+6) = 12/8 = 1.5 Ohm

I = V/R = 3/1.5 = 2 Amperes(the current through both the bulbs should sum to 2 ampere as a check)

Now we know that, the potential remains same for the resistors connected in parallel, So

V(C) = V(D) = 3 Volts

I(C) = V(C)/R(C) = 3/2 = 1.5 A

I(D) = V(D)/R(D) = 3/6 = 0.5 A (1.5+0.5=2A)

(d)Order of brightness:

D = B > A = C

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