a battery with and emf of 12V and an internal resistance of 0.5 ohms, is connected in series to a resistor of 5.0 ohms. what is the terminal voltage (n V) of the battery?
Completely inappropriate and only further insinuates your ignorance...You should be ashamed of yourself. Grow up and get a life those comments should be kept to yourself.
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To find the terminal voltage of the battery, we need to consider the concept of voltage drop due to internal resistance. The terminal voltage of the battery can be calculated using Ohm's Law.
Ohm's Law states that the voltage (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R). In this case, the resistor is the internal resistance of the battery.
The total resistance (R_total) in the circuit is the sum of the internal resistance (R_internal) and the resistance connected in series (R_resistor). So we have:
R_total = R_internal + R_resistor
Given:
EMF (E) = 12V
Internal resistance (R_internal) = 0.5 ohms
Resistance connected in series (R_resistor) = 5.0 ohms
Therefore, the total resistance (R_total) = 0.5 ohms + 5.0 ohms = 5.5 ohms
The current flowing through the circuit (I) can be calculated using Ohm's Law:
I = E / R_total
I = 12V / 5.5 ohms
Now that we have the current, we can calculate the voltage drop across the internal resistance. The voltage drop (V_internal) can be calculated using Ohm's Law:
V_internal = I * R_internal
Finally, the terminal voltage (V_terminal) of the battery can be calculated by subtracting the voltage drop across the internal resistance from the EMF:
V_terminal = E - V_internal
Now you can plug in the values and calculate the terminal voltage (V_terminal).