The discussion revolves around calculating the current in a series circuit and determining the voltage when a switch is closed. The original poster expresses confusion about how to find the voltage without knowing the current, despite having a specific answer in mind.
The discussion is active, with participants providing guidance and questioning assumptions about the circuit's parameters. Some participants have made calculations, while others are clarifying the implications of internal resistance and its effect on the voltage reading across the battery terminals.
There is mention of specific values for voltage and resistance, as well as the internal resistance of the battery, which may influence the calculations. The original poster's reference to a specific answer suggests a potential misunderstanding of the problem setup.
Doc Al said:You're given all the information needed to calculate the current.
No. What's the total resistance in the circuit?tweety1234 said:okay so would it be like this ;
6/ 10 = 0.6 Amps
Doc Al said:No. What's the total resistance in the circuit?
This problem asks for the voltmeter reading when the switched is closed.tweety1234 said:oh right, but i thought you don't include the internal resistance of the battery if the switch is open ?
That's the current.6/10.4 = 0.577
That's the voltage drop due to the internal resistance. But what's the full voltage reading across the battery terminals? (Does that 0.23 volts add or subtract from the battery's emf?)0577 x 0.4 = 0.23 volts