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Question

A cell has an internal resistance of 0.50 W. It has a terminal voltage of 1.4 V when connected to a 5.0 W external resistance. What will its terminal voltage be if the 5.0 W resistor is replaced by a 10.0 W resistor?



I need your explanation, of course.

If you need the terminal voltage formula, here it is: V=emf-Ir.
Please help me, I'm really stuck on this question.

Answer 1

Use the Thevenin equivalent to determine the internal unloaded source potential. (Draw a voltage source in series with the internal resistance of 0.5 ohms to the output terminal.) The current being drawn from the circuit in the initial condition is 1.4/5.0 = 0.28A. The drop across the internal resistance in this condition is 0.5*0.28 = 0.14V. Therefore the unloaded (internal) source voltage is 1.4 + 0.14 =1.54V. When a 10-ohm resistor is used as a load, the total current will be 1.54/(0.5+10), or 1.54/(10.5) = 0.147A. The terminal voltage is this current times the load resistance, or 10*0.147 = 1.47V

Answer 2

If E is the emf of the cell, a is the current in the first case and b is the current in the second case,

Total resistance in the first case is 5 + 0.5 = 5.5 ohm
Total resistance in the second case is 10 + 0.5 = 10 .5 ohm

E = 5.5 a = 10.5 b. ---------------1

a = 1.4 / 5 = 0.28 ( since 1.4 v is the voltage across 5 ohm.
b = x / 10,
x is the terminal voltage in the second case, voltage across 10 ohm.

Using equation 1,

10.5 (x / 10) = 5.5 * 0.28

x = 1.47V