Question

A battery with an emf of 12.0 V shows a terminal voltage of 11.4Vwhen operating in a circuit with two lightbulbs, each rated at 4.0 W(at 12.0 V),which are connected in parallel. What is the battery's internal resistance?

Express your answer to two significant figures and include the appropriate units.

Answer 1

0.95 A

Step-by-step explanation:

P = Power of bulb = 4 W

V = Voltage = 12 V

Power is given by

`P=(V^2)/(R)\\\Rightarrow R=(V^2)/(P)\\\Rightarrow R=(12^2)/(4)\\\Rightarrow R=36\ \Omega`

The connections are in Parallel

Equivalent resistance of these two bulb

`(36)/(2)=18\ \Omega`

Current from the source

`I=(11.4)/(18)=0.63\ A`

The voltage drop is

`12-11.4=0.6\ V`

Voltage drop is given by

`\Delta V=R_iI\\\Rightarrow R_i=(\Delta V)/(I)\\\Rightarrow R_i=(0.6)/(0.63)\\\Rightarrow R_i=0.952380952381\ A`

The internal resistance is 0.95 A