The circuit below contains three 100-watt light bulbs. The emf ε = 110 V. Which light bulb(s) is(are) the brightest?
A)A
B)B
C)C
D)B and C
E)All three are equally bright.

Question

The circuit below contains three 100-watt light bulbs. The emf ε = 110 V. Which light bulb(s) is(are) the brightest? 
A)A 
B)B 
C)C 
D)B and C 
E)All three are equally bright.

Quizplus · Accepted Answer

Assuming that the three light bulbs are identical, the brightness of the bulbs is determined by the current passing through each bulb. The circuit is a series circuit, so the same current passes through each bulb.

Using Ohm's law, we can calculate the current in the circuit:

I = ε / R

where ε is the emf and R is the total resistance of the circuit.

The resistance of each bulb is given by:

R = V^2 / P

where V is the voltage across the bulb (which is the same for all bulbs in this case) and P is the power of the bulb.

Substituting the given values, we get:

R = 110^2 / 100 = 121 Ω

The total resistance of the circuit is therefore:

R_total = 3R = 363 Ω

The current in the circuit is:

I = 110 / 363 = 0.303 A

Using Ohm's law again, we can calculate the voltage across each bulb:

V = IR

For bulb A, the voltage is:

V_A = 0.303 x 121 = 36.6 V

For bulb B, the voltage is:

V_B = V_A = 36.6 V

For bulb C, the voltage is:

V_C = V_A = 36.6 V

Finally, we can calculate the power dissipated by each bulb:

P = V^2 / R

For bulb A, the power is:

P_A = 36.6^2 / 121 = 11.1 W

For bulb B, the power is:

P_B = 36.6^2 / 121 = 11.1 W

For bulb C, the power is:

P_C = 36.6^2 / 121 = 11.1 W

Therefore, all three bulbs are equally bright.

The Circuit Below Contains Three 100-Watt Light Bulbs