Question

When we connect a number of light bulbs in a circuit with a battery, we can use our models of current, resistance, and Kirchhoff's principles to predict the relative brightness of the bulbs. If the bulbs are all identical, then the brightness of a bulb will be an indicator of the current through that bulb (i.e., a brighter bulb has more current).

1. The bulbs in the circuit shown on the right are all identical. Predict which bulb or bulbs will be the brightest, and which bulb or bulbs will be the dimmest. Also state if any bulbs will have equal brightnesses. Explain your prediction in detail.

All three bulbs have a resistance of 30 Ω, and the battery has a potential difference of 7 V.

2. Determine the current through each of the bulbs.
Current in bulb A =  A
Current in bulb B =  A
Current in bulb C =  A

3. From your answers to part 2, do you think your prediction in part 1 was correct? Why or why not?

4. Suppose that we disconnected bulb C, leaving just bulbs A and B. What would this do to the total resistance of all the bulbs connected to the battery, and would this cause bulb A to get brighter, dimmer, or stay the same? Explain your reasoning.

Answer 1

a)

• Bulb A will be brightest
• Bulb B and Bulb C will be dimmest with equal brightness.

b)

equivalent resistance

R_eq =30 + (1/30 + 1/30)^-1

R_eq = 45 ohms

Total current flowing in the circuit

I=V/R_eq = 7/45 =0.1556 A

Current flowing through each light bulb is

I_A =0.1556 A

I_B=I/2 =0.0778 A

I_C=I/2 =0.0778 A

c)

Yes ,since brightness of the bulb is given by

P=I²R

which means the greater the current flowing through the bulb,the brighter the bulb

d)

Total resistance

R=30+30 =60 ohms

Current flowing in the circuit

I=7/60 =0.1167 A

Current flowing through bulb A is

I_A=I_B=0.1167 A

Therefore Bulb A would be dimmer since lesser current is flowing compared to above .