A battery with an emf of 7.9 V and internal resistance of 1.37 Ω is connected across a load resistor

R.If the current in the circuit is 0.88 A, what is the value of R?
Answer in units of Ω

What power is dissipated in the internal resistance of the battery?
Answer in units of W

How do you set this problem up?

I=ℇ/(R+r)

R= (ℇ-Ir)/I= (7.9-0.88•1.37)/0.88 = …
P=I²r = …

i = 7.9/(1.37+R)

.88 = 7.9/(1.37+R)
1.37 + R = 8.98
so
R = 7.61 ohms

P = i^2 r = .88^2(1.37) = 1.06 Watts

To solve this problem, you can use Ohm's law and the concept of power dissipation.

1. Finding the value of R:
According to Ohm's law, the voltage across a resistor (V) is equal to the current (I) multiplied by the resistance (R):
V = I * R

In this case, the voltage across the load resistor (V) is the EMF of the battery which is 7.9 V, and the current (I) is given as 0.88 A. We need to find the value of R.

Hence, we can rearrange the formula to solve for R:
R = V / I

Substituting the given values:
R = 7.9 V / 0.88 A

Calculating the value of R gives us:
R ≈ 8.98 Ω

Therefore, the value of R is approximately 8.98 Ω.

2. Finding the power dissipated in the internal resistance:
The power dissipated in any resistor can be calculated using the formula:
P = I^2 * R

Here, we're looking to find the power dissipated in the internal resistance (P). The current (I) is given as 0.88 A, and the resistance of the internal resistor (R) is given as 1.37 Ω.

Substituting the given values:
P = (0.88 A)^2 * 1.37 Ω

Calculating the power dissipated in the internal resistance gives us:
P ≈ 1.01 W

Therefore, the power dissipated in the internal resistance of the battery is approximately 1.01 W.

To solve this problem, we can use Ohm's law and the concept of power.

For the first question, we can start by using Ohm's law, which states that the voltage across a resistor is equal to the current flowing through it multiplied by its resistance. In this case, the voltage across the load resistor (R) is the difference between the battery's electromotive force (emf) and the voltage drop across the internal resistance (r):

V = emf - Ir

Given that the emf is 7.9 V, the current is 0.88 A, and the internal resistance is 1.37 Ω, we can substitute these values into the equation:

V = 7.9 - 0.88 * 1.37

Simplifying the equation, we find:

V = 6.98 V

Since we know that V = IR (Ohm's law), we can rearrange the equation to solve for R, the resistance:

R = V / I = 6.98 / 0.88 = 7.93 Ω

So, the value of R is 7.93 Ω.

For the second question, we need to determine the power dissipated in the internal resistance of the battery. The power (P) dissipated in a resistor can be calculated using the formula:

P = I^2 * R

Given that the current (I) is 0.88 A and the resistance (R) of the internal resistance is 1.37 Ω, we can substitute these values into the equation:

P = (0.88)^2 * 1.37

Simplifying the equation, we find:

P = 1.0816 W

Therefore, the power dissipated in the internal resistance of the battery is 1.0816 W.