Question

A lightbulb is connected in series with a 7 Ω resistor and a 9.0 battery. If the current flowing through the circuit is 0.8 A, what is the internal resistance of the lightbulb (in ohms)? Round to the nearest tenth.

Please show work!!!!!

Answer 1

Using Ohm's Law, we can find the voltage across the circuit:

V = IR = (0.8 A)(7 Ω) = 5.6 V

The voltage across the lightbulb is:

VL = VB - VR = 9.0 V - 5.6 V = 3.4 V

Using Ohm's Law again, we can find the resistance of the lightbulb:

RL = VL / IL = 3.4 V / 0.8 A = 4.25 Ω

To find the internal resistance of the lightbulb, we subtract the resistance of the wires and the resistor from the total resistance of the circuit:

Rinternal = RL - Rwire - Rresistor

Assuming negligible resistance in the wires, we can simplify to:

Rinternal = RL - Rresistor = 4.25 Ω - 7 Ω = -2.75 Ω

The negative result means that the assumption of negligible resistance in the wires is not valid. There is some resistance in the wires, and the internal resistance of the lightbulb cannot be determined with the given information.