A battery has an internal resistance of 0.016 and an emf of 9.00 V. What is the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V?

Oh, I see we're talking about batteries now. Well, let's calculate that maximum current without making things too serious.

Okay, so you have a battery with an internal resistance and an emf. It's like a tough guy with a bit of resistance, you know? Anyway, the emf is 9.00 V, which is the battery's ideal voltage when there's no current flowing.

Now, you also don't want the terminal voltage to drop below 8.75 V. It's like the battery saying, "Hey, I don't wanna be too weak!"

To find the maximum current that can be drawn, we can use Ohm's Law. Remember, Ohm's Law is like a wise old electrician's advice. It says V = IR, where V is voltage, I is current, and R is resistance.

In this case, we have the terminal voltage (V) which is 8.75 V, and the internal resistance (R) which is 0.016 Ω. So we can rearrange the equation to solve for the maximum current (I).

I = (V - IR) / R

Alright, let's plug in the numbers and do some laughter... I mean calculations!

I = (8.75 V - (0.016 Ω * I)) / 0.016 Ω

Now, let's simplify this equation and solve for I. Oh wait, I forgot to tell you a funny joke!

Why did the math book look sad?

Because it had too many problems!

Alright, now let's solve for I, for real this time!

To find the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V, we can use Ohm's Law.

Ohm's Law states that V = I * R, where V is the voltage across the circuit, I is the current flowing through the circuit, and R is the resistance in the circuit.

In this case, the voltage across the circuit is the terminal voltage, which is 8.75 V. The resistance in the circuit is the sum of the internal resistance of the battery and any external resistance in the circuit.

R = internal resistance + external resistance = 0.016 + 0 = 0.016

Now, we can rearrange Ohm's Law to solve for the maximum current:

I = V / R

I = 8.75 V / 0.016 Ω

Calculating this:

I = 546.875 A

Therefore, the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V is approximately 546.875 A.

To find the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V, we need to consider the voltage drop across the internal resistance.

The voltage drop across the internal resistance can be calculated using Ohm's Law: V = IR, where V is the voltage drop, I is the current, and R is the resistance.

In this case, the voltage drop across the internal resistance can be found by subtracting the terminal voltage (8.75 V) from the electromotive force (EMF) of the battery (9.00 V):

V_drop = EMF - Terminal Voltage
V_drop = 9.00 V - 8.75 V
V_drop = 0.25 V

Since the internal resistance of the battery is given as 0.016 Ω, we can now use Ohm's Law to find the maximum current:

V_drop = IR
0.25 V = I * 0.016 Ω

To solve for I, we rearrange the equation:

I = V_drop / R
I = 0.25 V / 0.016 Ω
I ≈ 15.63 A

Therefore, the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V is approximately 15.63 Amperes.

Ri = 0.016 Ohms.

E = 9.00 Volts.

Imax = V/Ri = (9.00-8.75) / 0.016 = 15.625 Amps.