Potential in an open circuit at the terminals.?

Question

When a 100 ohm resistor is connected to a battery, the terminal voltage is 30.6 V. When a 50 ohm resistor is connected the value is 29.4 V. What would the potential be with an open circuit at the terminals?

Answer 1

♦ I=E/(R+r); r is battery’s resistance;
I1=E/(100+r)= 30.6/100; hence
100+r =E*100/30.6;
I2=E/(50+r)= 29.4/50; hence
50+r = E*50/29.4;
♥ (100+r) –(50+r) =E*(100/30.6 -50/29.4);
♠ E= 50/(100/30.6 -50/29.4) =31.9 V;

Answer 2

Simple.
The short answer is you should see the same voltage across the resistor as you see with the open terminals provided there has been no voltage drop due to drain and usage.

The difference is how much voltage was used up by the first resistor lowering the load voltage. This is only a 3% difference and within the range of a drained battery. If you recharge the battery the voltage will be the same for both resistors. And if you are using a resistor with too low a wattage rating, the heat loss will quickly lower the battery voltage. You should use more resistance for this much voltage or lower the voltage.
Check the battery voltage without a resistor in circuit.
You will see the same voltage across a resistance network of any kind provided you have enough ohms not to be a short to the voltage source. I hope your resistors are of high enough wattage for your experiment. If they get really hot they are not rated for your circuit.

Look at the current.

E = I x R

I = E / R = 30.6volts / 100 ohms

I = .306 amps

Now check the power in wattage for the circuit.

P = E x I = 30.6 volts x .306 amps

P = 9.364 watts.

You must use a resistor rated for at least 10 watts.

Finally the battery internal resistance will be the same for both resistors so it does not effect the voltage drop difference between the two. I would bet a nickel you are using a much lower watt rated resistor.

Play safe.

Answer 3

Assume a standard single-resistor model for your battery (voltage source + resistor in series)
V0 will be the potential of -only- the voltage source (and thus the open-circuit voltage)
Measured voltage drop accross the main resistor will be:
V0 * (R_main/(R_main + R_battery))
Solve the system of equations:
V0*(100/(100+R_batt)) = 30.6
V0*(50/(50+R_batt)) = 29.4
Solution:
V0 = 31.9 V <== Answer
R_batt = 4.255 Ohm

Answer 4

Roughly around 32 volts. if you neglect the internal resistance of the battery. But to be accurate you need to know the internal resistance of the battery.

Answer 5

Zero.

Answer 6

infinity. oops thoought you were asking resistance