A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.7 A when an additional 3 Ω resistor is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero.

I got 12.75 however that isn't right

Volts = V = iR = 2.2 R1

so
2.2 R1 = 1.7 (R1+3)
2.2 R1 = 1.7 R1 + 5.1
.5 R1 = 5.1
R1 = 10.2

V2 = I*R2 = 1.7 * 3 = 5.1 Volts.

R1 = Change in voltage/change in current = -5.1/(1.7-2.2) = 10.2 Ohms.

To find the value of R1, we can use Ohm's Law and Kirchhoff's Voltage Law (KVL).

Let's start by identifying the given information:
Resistance before adding the additional resistor: R1
Current before adding the additional resistor: 2.2 A
Current after adding the additional resistor: 1.7 A
Resistance of the additional resistor: 3 Ω

According to Ohm's Law, the voltage across a resistor is equal to the current flowing through it multiplied by its resistance (V = IR).

Using this formula, we can calculate the voltage across the circuit before and after adding the additional resistor.

Voltage across the circuit before adding the additional resistor:
V1 = 2.2 A * R1

Voltage across the circuit after adding the additional resistor:
V2 = 1.7 A * (R1 + 3 Ω)

According to Kirchhoff's Voltage Law (KVL), the sum of the voltage drops across the resistors in a closed loop circuit is equal to the voltage supplied by the source.

Therefore, V1 must be equal to V2:
2.2 A * R1 = 1.7 A * (R1 + 3 Ω)

Now, let's solve this equation to find the value of R1.

2.2 A * R1 = 1.7 A * R1 + 1.7 A * 3 Ω
2.2 A * R1 - 1.7 A * R1 = 1.7 A * 3 Ω
0.5 A * R1 = 5.1 Ω
R1 = 5.1 Ω / 0.5 A
R1 = 10.2 Ω

Therefore, the value of R1 is 10.2 Ω.

To find the value of R1, we can use Ohm's Law and Kirchhoff's Voltage Law (KVL).

Let's start by applying Ohm's Law:
V = I * R

Where:
V is the voltage across the resistor
I is the current passing through the resistor
R is the resistance of the resistor

In the first scenario, when only R1 is present, we have:
V1 = I1 * R1 ----- Equation 1

Given that the current passing through R1 is 2.2 A, we can rewrite Equation 1 as:
V1 = 2.2 * R1 ----- Equation 2

In the second scenario, when the 3 Ω resistor is added in series with R1, the current passing through the circuit is reduced to 1.7 A. Therefore, the voltage across R1 becomes:
V2 = 1.7 * R1 ----- Equation 3

Since the two resistors are in series, the total resistance of the circuit becomes:
R_total = R1 + 3

Now, according to Kirchhoff's Voltage Law (KVL), the sum of voltage drops across the resistors should equal the voltage supplied by the power source (assuming the internal resistance of the source of emf is zero). Therefore, we have:
V1 + 3I1 = V2 ----- Equation 4

Substituting the values from Equations 2 and 3 into Equation 4, we get:
2.2 * R1 + 3 * 2.2 = 1.7 * R1

Expanding the equation:
2.2 * R1 + 6.6 = 1.7 * R1

Rearranging the terms:
2.2 * R1 - 1.7 * R1 = 6.6
0.5 * R1 = 6.6

Solving for R1:
R1 = 6.6 / 0.5
R1 = 13.2 Ω

Therefore, the value of R1 is 13.2 Ω, not 12.75 Ω as you calculated.