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Calculate the terminal voltage for a battery with an internal resistance of 0.675 Ω and an emf of 6.50 V when the battery is connected in series with an 85.0 Ω resistor.

Calculate the terminal voltage for a battery with an internal resistance of 0.675 Ω and an emf of 6.50 V when the battery is connected in series with an 850 Ω resistor.

User Kramer Li
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Final answer:

The terminal voltage of a battery can be calculated using the equation V = emf - Ir. The terminal voltage for a battery with an internal resistance of 0.675 Ω and an emf of 6.50 V, when connected in series with an 85.0 Ω resistor, is 6.452 V. When connected in series with an 850 Ω resistor, the terminal voltage is 6.494 V.

Step-by-step explanation:

The terminal voltage of a battery can be calculated using the equation V = emf - Ir, where V is the terminal voltage, emf is the electromotive force or emf of the battery, I is the current flowing through the circuit, and r is the internal resistance of the battery.

In this case, the emf is given as 6.50 V and the internal resistance is 0.675 Ω. When the battery is connected in series with an 85.0 Ω resistor, the current flowing through the circuit can be found using Ohm's Law (I = V/R) which gives:

I = 6.50 V / (85.0 Ω + 0.675 Ω) = 0.0752 A.

Substituting the values into the terminal voltage equation, V = 6.50 V - (0.0752 A * 0.675 Ω) = 6.452 V.

When the battery is connected in series with an 850 Ω resistor, the current flowing through the circuit can be found in the same way:

I = 6.50 V / (850 Ω + 0.675 Ω) = 0.0075 A.

Substituting the values, V = 6.50 V - (0.0075 A * 0.675 Ω) = 6.494 V.

User Pengguna
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