Question

Calculate the terminal voltage for a battery with an internal resistance of 0.900 Ω and an EMF of 6.00 V when the battery is connected in series with (a) a 71.0-Ω resistor, and (b) a 710-Ω resistor.

a) (a) 4.24 V, (b) 4.86 V
b) (a) 5.63 V, (b) 5.98 V
c) (a) 3.72 V, (b) 3.91 V
d) (a) 7.15 V, (b) 7.50 V

Answer 1

The terminal voltage for a battery can be calculated using the equation V = emf - Ir, where V is the terminal voltage, emf is the electromotive force of the battery, I is the current flowing through the circuit, and r is the internal resistance of the battery. Using this formula, we can determine the terminal voltage for the given battery and resistors.

Step-by-step explanation:

When a battery is connected to an external load, the terminal voltage can be calculated using the equation V = emf - Ir, where V is the terminal voltage, emf is the electromotive force or the voltage provided by the battery, I is the current flowing through the circuit, and r is the internal resistance of the battery.

In the given question, we have an EMF of 6.00 V and an internal resistance of 0.900 Ω. When connected in series with a 71.0-Ω resistor, the terminal voltage can be calculated as follows:

V = emf - Ir = 6.00 V - (I * 0.900 Ω) = 6.00 V - (I * 71.0 Ω)

To find the current, we can use Ohm's Law: I = V / R. Substituting the values, we get:

I = (6.00 V) / (71.0 Ω + 0.900 Ω) = 6.00 V / 71.9 Ω = 0.0834 A

Now, substituting this value of I into the equation for terminal voltage, we get:

V = 6.00 V - (0.0834 A * 71.0 Ω) = 6.00 V - 5.908 Ω = 0.092 V

Therefore, the terminal voltage for a battery with an internal resistance of 0.900 Ω and an EMF of 6.00 V when connected in series with a 71.0-Ω resistor is approximately 0.092 V. Similarly, for a 710-Ω resistor, the terminal voltage can be calculated in the same way.