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?
R + 11 = (11A/6.6A)R 1.666R
1.666R - 1R = 11
0.666R = 11
R = 16.5 Ohms. = Original Resistance.
Correction: R + 11=(11A/6.6A)*R= 1.666R.
To find the resistance in the original circuit, we can use Ohm's Law and the equation for calculating the total resistance in a series circuit.
Ohm's Law states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R), represented as I = V/R.
In a series circuit, the total resistance (RT) is the sum of the individual resistances. So, RT = R1 + R2 + R3 + ...
Given that the current in the original circuit is 11.0 A, and the current drops to 6.6 A when an additional 11.0-Ω resistor is inserted, we can set up the following equations:
For the original circuit:
11.0 A = V/RT (Equation 1)
For the circuit with the additional resistor:
6.6 A = V/(RT + 11.0 Ω) (Equation 2)
Since we're only interested in finding the resistance in the original circuit, we can rearrange Equation 1 to solve for RT:
RT = V/11.0 A (Equation 3)
Substituting Equation 3 into Equation 2, we can solve for V (voltage):
6.6 A = V/[V/11.0 A + 11.0 Ω]
Multiply both sides of the equation by V/11.0 A + 11.0 Ω:
6.6 A * (V/11.0 A + 11.0 Ω) = V
Distribute the 6.6 A:
(6.6 A * V)/11.0 A + 6.6 A * 11.0 Ω = V
Cancel out the A's:
(6.6 V)/11.0 + 72.6 Ω = V
Multiply both sides of the equation by 11.0 to eliminate the fraction:
6.6 V + 798.6 Ω = 11.0 V
Rearrange the equation to solve for V:
4.4 V = 798.6 Ω
V = 798.6 Ω / 4.4
V ≈ 181.5 Ω
Now, substitute the value of V back into Equation 3 to find RT:
RT = 181.5 Ω / 11.0 A
RT ≈ 16.5 Ω
Therefore, the resistance in the original circuit is approximately 16.5 Ω.