Question

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 :(

Answer 1

V = IR
11.7R = 6.83(R+8.87)
R = 12.434

Answer 2

can you write this out in a few more steps/equations i can't follow what you have done

Answer 3

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.