Question

A. Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance of each light bulb remains constant. Rank the bulbs (A through E) based on their brightness.

Rank from brightest to dimmest. To rank items as equivalent, overlap them.

B. What happens to bulb C?

• It gets dimmer.

• It gets brighter.

• There is no change.

• 
  

Answer 1

Concepts and reason

The concepts used to solve this problem are effective resistance of the combination of resistor, potential across the bulbs and the power dissipated by the bulbs.

Initially, simplify the circuit by using a parallel and series combination of resistors. Next, rank the bulb based on brightness. Later, simplify the circuit when the switch is open and rank the bulb based on brightness.

Fundamentals

The expression for a series combination of resistors is,

Here, is the equivalent resistance of bulbs in series, is the resistance of one bulb, and is the resistance of the other bulb.

The expression for a parallel combination of resistors is,

Here, is the equivalent resistance of bulbs in a parallel combination.

The expression for power in the circuit is,

Here, P is the power, I is the current, and R is the resistance.

The expression for current in the circuit is,

Here, I is the current in the circuit, V is the potential, and R is the resistance

(A)

Bulb A and B are connected in parallel.

The expression to find the effective resistance for the parallel connection is,

Here, is the effective combination of bulbs AB connected in parallel, and are the resistance of the bulb A and B.

Substitute R for and R for .

Bulb D and E are connected in the series.

The expression for a series combination of resistors is,

Here, is the effective combination of bulbs DE connected in series, and are the resistance of bulb D and E.

Substitute R for and R for .

The bulb DE is parallel with bulb C. So, the effective resistance of theirs is,

Substitute R for 2R for .

The expression for current in the circuit is,

Here, is the current through the bulb A and B, V is the potential, and is the resistance of the bulb AB.

Substitute for .

Since A and B have equal resistance and they are connected in parallel, the current is divided equally through them; that is,

and .

The expression for current in the circuit is,

Here, is the current through the bulb D and E and is the resistance of the bulbs DE.

Substitute for R

The expression for current in the circuit is,

Here, is the current through the bulb C, D and E and is the resistance of the bulbs C, D and E.

Substitute for .

The expression for the power dissipated by the bulb is,

Here, P is the power dissipated by the bulb.

The expression for the power dissipated by the bulb in the circuit A is,

Here, is the power dissipated by the bulb A.

Substitute for and R for to find .

Since the bulbs A and B have same resistance and the same current pass through them.

They dissipate equal power.

Here, is the power dissipated by the bulb B.

Substitute for in the above expression.

The expression for the power dissipated by the bulb in the circuit C is,

Since there are three bulbs in the circuit, bulb C is in parallel with bulb DE where 2/3 of current passes through bulb C and 1/3 of current passes through bulb DE, since

and .

Use for to rewrite the equation .

Substitute for and R for to find .

Since the bulb D and E are in series, the power dissipated by them are the same.

Here, is the power dissipated by the bulb D and is the power dissipated by the bulb E.

The expression for the power dissipated by the bulb in the circuit C is,

Substitute for and R for to find .

Since the bulb D and E are in series, the power dissipated by them are same.

Substitute for in the above expression to find .

(B)

Opening the switch increases the resistance of the circuit.

This is because the effective resistance is less than . Due to an increase in resistance, the current through bulb A is reduced.

Ans: Part A

The rank of power in the bulb based on brightness is.

Part B

The brightness of bulb A is dimmed.

Part C

Bulb C is brighter because it has more current.