Question

(A) For circuit 2 above, rank the bulbs in order of brightness. ( use symbols <,>,= )

(B) For circuit 3 above, rank the bulbs in order of brightness. ( use symbols <,>,= )

(C) if bulb B is removed from circuit 2, what will happen to bulbs A and C? (Choose all that apply)

•A stays the same

•A goes out

•A gets brighter

•A gets dimmer

•C stays the same

• C goes out

• C gets brighter

• C gets dimmer

(D) if bulb B is removed from circuit 3, what will happen to bulbs A and C? ( choose all that apply)

•A stays the same

•A goes out

•A gets brighter

•A gets dimmer

•C stays the same

• C goes out

• C gets brighter

• C gets dimmer

(E) if bulb B is removed from circuit 4, what will happen to bulbs A and C? ( choose all that apply)

•A stays the same

•A goes out

•A gets brighter

•A gets dimmer

•C stays the same

• C goes out

• C gets brighter

• C gets dimmer

Answer 1

Concepts and reason

The concepts used to solve this problem are series and parallel combination of resistors and Ohm’s law.

Initially, using Ohm’s law rank the bulbs in order of brightness for given circuits 2 and 3. After that using the parallel combination of resistance in circuit 2, explain what will happened to the bulbs A and C if the bulb B is removed from the circuit.

Similarly, using the parallel and series combination of resistors concept explain what will happened to the bulbs A and C if the bulb B is removed from both the circuits 3 and 4 respectively.

Fundamentals

Ohm’s law is defined as the current in an electric circuit is directly proportional to the applied voltage and inversely proportional to the resistance and t is expressed as:

$\begin{array}{c}\\I = \frac{V}{R}\\\\V = IR\\\end{array}$

Here, I is the current, V is the voltage, and R is the resistance.

If ${R_1}$, ${R_2}$, and ${R_3}$ are the three resistors in an electrical circuit, each resistor in a series circuit has the same amount of current flowing through it. And current through each resistor in a parallel combination is different, which depends up on individual resistances.

(A)

Ohm’s law is defined as the current in an electric circuit is directly proportional to the applied voltage and inversely proportional to the resistance and t is expressed as:

$\begin{array}{c}\\I = \frac{V}{R}\\\\V = IR\\\end{array}$

From the given circuit 2, the three bulbs are connected in parallel combination and its resistance is same for the three bulbs. Therefore, the current passing through each bulb is same.

Therefore, the rank of the bulbs in order of brightness is $A = B = C$.

(B)

From circuit 3, the bulbs B and C are connected in parallel combination and the bulb A is connected in series combination. So, the current passes through the bulbs C and B is same and is less than A. therefore, the current passes through the bulb A is more when compared to the bulbs B and C in the given circuit 3.

Therefore, the rank of the bulbs in order of brightness is$A > B = C$.

(C)

From the given circuit 2, the three bulbs are connected in parallel combination and its resistance is same for the three bulbs. Therefore, the current passing through each bulb is same.

Since, the current passing through each bulb is same, therefore the brightness of bulbs B and C are also same even after bulb B is removed from the circuit. Therefore, the bulbs A and C stays same in circuit 2.

(D)

Now, if the bulb B is removed then the total resistance of the given circuit increases with decrease in total current coming from the battery.

From circuit 3, the bulbs B and C are connected in parallel combination and the bulb A is connected in series combination with battery. Then there is low current passing through the bulb A here. And for bulb C from parallel combination of bulbs B and C, the current is now going to increase since, the current is not going to divided among the resistors B and C (since B is already removed). Hence, the bulb C gets brighter.

(E)

If the bulb B is removed from circuit 4, here, bulb A is connected in parallel combination with the battery, then there is no change in voltage across bulb A after removing the bulb B. therefore, the brightness of the bulb A stays the same.

Finally, if the bulb B is removed from the series combination of bulbs B and C the voltage across bulb C increases therefore, the brightness of the bulb C is also increases.

Ans: Part A

The required rank of the bulbs in order of brightness is $A = B = C$.

Part B

The required rank of the bulbs in order of brightness is $A > B = C$.

Part C

The required bulbs A and C stays same in circuit 2.

Part D

The required bulbs A gets dimmer and C gets brighter in circuit 3.

Part E

The required bulbs A stays same and C gets brighter in circuit 4.