3) What is the emf of a battery with an internal resistance of 0.15 if the battery delivers 1.5 A to an externally connected 5.0 resistors?
Ohm my goodness, that's a shocking question! To find the emf of the battery, we need to use Ohm's law, which is a real watt of a time. The equation we're after is:
emf = (current + internal resistance) × external resistance
Given that the current is 1.5A, the internal resistance is 0.15Ω, and the external resistance is 5.0Ω, we can plug those values in:
emf = (1.5A + 0.15Ω) × 5.0Ω
Now, let's do the math, but be careful not to get too amped up:
emf = (1.65) × 5.0Ω
emf = 8.25V
So the emf of this electrifying battery is a shocking 8.25 volts! Don't let it zap you!
To find the emf of the battery in this scenario, you can use Ohm's Law along with the concept of internal resistance.
Step 1: Calculate the voltage drop across the external resistance.
Using Ohm's Law, the voltage drop across the external resistance (V) can be calculated using the formula:
V = I * R,
where I is the current flowing through the resistance and R is the resistance.
In this case, the current flowing through the resistance is given as 1.5 A and the resistance is given as 5.0 Ω.
Therefore, V = 1.5 A * 5.0 Ω = 7.5 V.
Step 2: Calculate the voltage drop across the internal resistance.
The voltage drop across the internal resistance (V_internal) can be calculated using Ohm's Law:
V_internal = I * r,
where r is the internal resistance of the battery.
In this case, the internal resistance is given as 0.15 Ω and the current flowing through the external resistance is 1.5 A.
Therefore, V_internal = 1.5 A * 0.15 Ω = 0.225 V.
Step 3: Calculate the emf of the battery.
The emf of the battery (E) is the voltage delivered by the battery when no current is flowing through the internal resistance. It is equal to the sum of the voltage drops across the external resistance and internal resistance.
Therefore, E = V + V_internal = 7.5 V + 0.225 V = 7.725 V.
So, the emf of the battery with an internal resistance of 0.15 Ω is 7.725 V.
To find the electromotive force (emf) of a battery with an internal resistance, you can use Ohm's Law and the concept of potential difference.
First, let's understand the given information:
- The internal resistance of the battery is 0.15 Ω.
- The battery delivers a current of 1.5 A to a 5.0 Ω resistor.
Now, let's apply Ohm's Law to the external circuit:
V = IR
Where:
V is the potential difference (voltage) across the external resistor,
I is the current flowing through the circuit, and
R is the resistance of the external resistor.
From the given information, we have:
I = 1.5 A (current)
R = 5.0 Ω (resistance)
Substituting these values into Ohm's Law, we can find the potential difference across the external resistor:
V = (1.5 A) * (5.0 Ω)
V = 7.5 V
Now, let's calculate the emf of the battery:
The emf (ε) of the battery is the maximum potential difference it can provide when there is no current flowing through the internal resistance. It can be found using the equation:
ε = V + Ir
Where:
ε is the emf of the battery,
V is the potential difference across the external resistor,
I is the current flowing through the circuit, and
r is the internal resistance of the battery.
From the given information, we have:
V = 7.5 V (potential difference across the external resistor)
I = 1.5 A (current)
r = 0.15 Ω (internal resistance)
Substituting these values into the emf equation, we can find the emf of the battery:
ε = 7.5 V + (1.5 A) * (0.15 Ω)
ε = 7.5 V + 0.225 V
ε = 7.725 V
Therefore, the emf of the battery is 7.725 volts.
voltage across resistor = i R = 1.5*5 = 7.5 Volts
voltage loss to internal resistance = 1.5 * .15 = .225 volts
total voltage = 7.5 + .225 = 7.725 Volts