Question

The three bulbs in (Figure 1) are identical.

Part A Rank the bulbs from brightest to dimmest. Rank the bulbs from brightest to dimmest. To rank items as equivalent, overlap them.

Part B

Suppose a wire is connected between points 1 and 2. What happens to bulb A?

Part C

What happens to bulb B?

Part D

What happens to bulb C?

Answer 1

Concepts and reason

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

Initially, simplify the circuit by using a parallel and series combination of resistors. Later, rank the bulb based on brightness by using the relation between the power and the current.

Finally, use the relation between the

Fundamentals

The expression for a parallel combination of resistors is,

Here, is the equivalent resistance of bulbs in a parallel combination, is the resistance of one bulb, and is the resistance of the other bulb.

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)

Consider the resistance of each bulb as R.

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

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

Substitute R for and R for .

The combination of the resistors B and C and the resistor A are connected in series.

So, the equivalent resistance of the resistors connected in series is given by,

Here, is the resistance of the bulb A.

Substitute for and R for in equation as follows:

The current passing through the circuit is given by,

Here, is the current through the circuit, V is the potential across the battery, and is the resistance of the circuit.

Substitute for in equation as follows:

The bulbs A and the combination of bulbs B and C are connected in series. So, the current passing through the circuit the bulb A and the combination of bulbs B and C are same as the current passing through the circuit.

The current across the bulb A is given by,

The power across the bulb A is given by,

Substitute for in equation as follows:

The voltage across the combination of bulbs B and C is given by,

Substitute for I and for in equation as follows:

The bulbs B and C are connected in parallel. So, the voltage across the bulbs B and C is equal to the voltage across the combination of bulbs.

The expression for current in the bulb B is,

Substitute for and R for in equation as follows:

The power across the bulb B is given by,

Substitute for and R for in equation as follows:

The expression for current in the bulb C is,

Substitute for and R for in equation as follows:

The power across the bulb C is given by,

Substitute for and R for in equation as follows:

The power across the bulb B and C is equal and is less than the power across the bulb A.

(B)

When the wire is connected across points 1 and 2 then the Bulbs B and C gets shorted.

Therefore, the current across the bulb A is given by,

(C)

When the wire is connected across the points 1 and 2 there is no current passing through the bulb B.

The expression for the power across the bulb B is given by,

From the above expression, the power across the bulb is directly proportional to the square of current across the bulb.

So, the power across the bulb B is zero.

(D)

When the wire is connected across the points 1 and 2 there is no current passing through the bulb C.

The expression for the power across the bulb C is given by,

From the above expression, the power across the bulb is directly proportional to the square of current across the bulb.

So, the power across the bulb C is zero.

Ans: Part A

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

Part B

The brightness of bulb A is increases.

Part C

The bulb B gets shorted.

Part D

The bulb C gets shorted.