The current in a series circuit is 11.7 A. When an additional 8.87-Ω resistor is inserted in series, the current drops to 6.83 A. What is the resistance in the original circuit?
I tried V=I(Ro+R) then with found voltage (60.61758V) used R=V/I but my answer was incorrect :(
V = IR
11.7R = 6.83(R+8.87)
R = 12.434
can you write this out in a few more steps/equations i can't follow what you have done
To find the resistance in the original circuit, we can use Ohm's Law, which states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R).
Let's start by calculating the initial resistance (Ro) of the circuit using the given information. We know that the initial current (I1) is 11.7 A and that when an additional resistor is inserted, the current (I2) drops to 6.83 A.
Using Ohm's Law, we can write:
I1 = V / Ro (equation 1)
I2 = V / (Ro + 8.87) (equation 2)
We also know that the voltage (V) in the circuit is constant.
Now, we can isolate V in equation 1:
V = I1 * Ro
Substituting this value of V into equation 2, we get:
I2 = (I1 * Ro) / (Ro + 8.87)
Next, we can solve equation 2 for Ro. Rearranging the equation, we have:
Ro + 8.87 = (I1 * Ro) / I2
To solve for Ro, we cross multiply:
Ro * I2 + 8.87 * I2 = I1 * Ro
Expand and simplify the equation:
Ro * I2 + 8.87 * I2 = I1 * Ro
Ro * I2 - I1 * Ro = -8.87 * I2
Factor out Ro:
Ro * (I2 - I1) = -8.87 * I2
Finally, divide both sides by (I2 - I1) to solve for Ro:
Ro = (-8.87 * I2) / (I2 - I1)
Plugging in the given values: I1 = 11.7 A, I2 = 6.83 A, and Ro = resistance of the original circuit, we can calculate Ro:
Ro = (-8.87 * 6.83) / (6.83 - 11.7)
Ro = (-60.4921) / (-4.87)
Ro ≈ 12.43 Ω
Therefore, the resistance in the original circuit is approximately 12.43 Ω.
Eq1: R = E/11.7 = Resistance of original circuit.
Eq2: R+8.87 = E/6.83 = Resistance of circuit with 8.87 ohms added.
In Eq2, replace R with E/11.7 and solve for E:
E/11.7 + 8.87 = E/6.83,
0.0855E + 8.87 = 0.1464E,
E = 145.6 volts = supply voltage.
R = E/I = 145.6/11.7 = 12.4 Ohms.