or is one of my processes not right.
http://k.minus.com/jbm3bGtvCAn3Ca.jpg
http://k.minus.com/jbez7eVzMxMuYb.jpg
I found the total current to be 0.40 A, and if I subtracted the 0.20 A given in the lamp, the current running through the other side should also be 0.20 A. But when i solved for current by adding the internal resistance of the battery by the resistance I got 0.30 A. So I don't know which current is right.
And the emf question, is emf the voltage it has before it goes through the internal resistance :x? I'm so confused now.
For the graph question, do u find emf by extending the slope backwards until you reach a y-int and internal resistance by finding the slope of the line, and would the negative of the slope be ignored? Sorry for asking so much questions, thanks in advance.
http://k.minus.com/jbm3bGtvCAn3Ca.jpg
http://k.minus.com/jbez7eVzMxMuYb.jpg
I found the total current to be 0.40 A, and if I subtracted the 0.20 A given in the lamp, the current running through the other side should also be 0.20 A. But when i solved for current by adding the internal resistance of the battery by the resistance I got 0.30 A. So I don't know which current is right.
And the emf question, is emf the voltage it has before it goes through the internal resistance :x? I'm so confused now.
For the graph question, do u find emf by extending the slope backwards until you reach a y-int and internal resistance by finding the slope of the line, and would the negative of the slope be ignored? Sorry for asking so much questions, thanks in advance.
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No, wait. You found the current through the resistor to be .40 A. Total current is .40 + .20 = .60 so total resistance is 6V/.6A = 10 ohms.
Total current flows through the battery developing .60 x 5 = 3.0 V. That is in series with the 6V across the circuit, so battery voltage is 9V.
To use the graph you extend the line to 0 A and 4.5 V and that is open circuit voltage. Extend the line the other way to 0 V and 3.75 A. Then 4.5/3.75 = 1.2 ohms.
Total current flows through the battery developing .60 x 5 = 3.0 V. That is in series with the 6V across the circuit, so battery voltage is 9V.
To use the graph you extend the line to 0 A and 4.5 V and that is open circuit voltage. Extend the line the other way to 0 V and 3.75 A. Then 4.5/3.75 = 1.2 ohms.