Question

A 16-V battery powers a series circuit that contains two lightbulbs The

voltage across the first lightbulb is 12-V​

Answer 1

Complete question:

A 16-V battery powers a series circuit that contains two lightbulbs. The voltage across the first lightbulb is 12-V.

a. What fractional part of the circuit's total energy is transformed in the first lightbulb? Simply your answer.

b. What fractional part of the circuit's total energy is transformed in the second lightbulb? Simply your answer.

Answer:

(a) Fraction of energy of the first bulb in the total energy of circuit is 3/4

(b) Fraction of energy of the second bulb in the total energy of circuit is 1/4

Step-by-step explanation:

Given;

total voltage across the series circuit,
`V_T` = 16 V

voltage across the first light bulb, V₁ = 12 V

The voltage across the second light bulb is given by;

V₂ =
`V_T` - V₁

V₂ = 16 V - 12 V

V₂ = 4 V

The total power across the circuit is given by;

P = I
`V_T`

P = 16I

Where;

I is the constant current in the circuit

(a)

power in the first light bulb is given by;

P₁ = 12I

Fraction of this power in the total energy of circuit is given by;

`(P_1)/(P) = (12I)/(16I) = (3)/(4)`

(b)

power in the second light bulb is given by;

P₂ = 4I

Fraction of this power in the total energy of circuit is given by;

`(P_2)/(P) = (4I)/(16I) = (1)/(4)`