The current in a series circuit is 11.0 A. When an additional 11.0-Ω resistor is inserted in series, the current drops to 6.6 A. What is the resistance in the original circuit?

E/R = 11.0

E/(R+11) = 6

Since the voltage drop is the same in both cases,

11.0R = 6.6(R+11)
R = 16.5Ω

To find the resistance in the original circuit, we can use Ohm's Law and apply the concept of series circuits.

In a series circuit, the total resistance is the sum of individual resistances. Let's denote the resistance in the original circuit as R1.

According to Ohm's Law, the current in a circuit (I) is equal to the voltage (V) divided by the resistance (R). Using this formula, we can set up two equations based on the given information:

Equation 1: I1 = V / R1
Equation 2: I2 = V / (R1 + 11.0)

Where:
I1 = initial current (11.0 A)
I2 = current after adding the additional resistor (6.6 A)
R1 = original resistance (unknown)
11.0 = resistance of the additional resistor (Ω)

Since we have two equations and two unknowns (V and R1), we can solve for R1.

First, rearrange Equation 1 to solve for V:
V = I1 * R1

Now, substitute this value of V into Equation 2:
I2 = (I1 * R1) / (R1 + 11.0)

Multiply through by (R1 + 11.0) to eliminate the denominator:
I2 * (R1 + 11.0) = I1 * R1

Expand:
I2 * R1 + 11.0 * I2 = I1 * R1

Rearrange the equation to isolate R1:
I2 * R1 - I1 * R1 = -11.0 * I2

Factor out R1:
R1 * (I2 - I1) = -11.0 * I2

Divide both sides by (I2 - I1) to solve for R1:
R1 = (-11.0 * I2) / (I2 - I1)

Substitute the given values into the equation:
R1 = (-11.0 * 6.6) / (6.6 - 11.0)

Calculating this expression gives us:
R1 ≈ 39.6 Ω

Therefore, the resistance in the original circuit is approximately 39.6 ohms.