Question

Light bulbs of fixed resistance 3.0 Ω and 6.0 Ω, a 9.0 V battery, and a switch S are connected as shown in the schematic diagram above. The switch S is closed.

a. Calculate the current in bulb A.

b. Which light bulb is brightest? Justify your answer.

c. Switch S is then opened. By checking the appropriate spaces below, indicate whether the brightness of each light bulb increases, decreases, or remains the same. Explain your reasoning for each light bulb.

i. Bulb A: The brightness .............. increases ..................... decreases ................ remains the same Explanation:

ii. Bulb B: The brightness ............... increases ................. decreases ................ remains the same Explanation:

iii. Bulb C: The brightness .................. increases ................decreases ............. remains the same Explanation:

Answer 1

a. The resistance of the 6 Ω and 3 Ω resistors in parallel is (6 × 3)/(6 + 3) = 2 Ω. Adding the 3 Ω resistor in the main branch gives a total circuit resistance of 5 Ω. The current in bulb A in the main branch is the total current delivered by the battery I = ε/R = (9 V)/(5 Ω) = 1.8 A

b. Bulb A is the brightest. In the main branch, it receives the most current. You can also calculate the power of each resistor where P_A = 9.7 W, P_B = 2.2 W and P_C = 4.3 W

c. i. Removing Bulb C from the circuit changes the circuit to a series circuit, increasing the total resistance and decreasing the total current. With the total current decreased, bulb A is dimmer.

ii. Since bulb A receives less current, the potential drop is less than the original value and being in a loop with bulb B causes the voltage of bulb B to increase, making bulb B brighter. The current through bulb B is greater since it is no longer sharing current with bulb C.

iii. The current through bulb C is zero, bulb C goes out.