Question

The current in a series circuit is 24.0 A. When an additional 25.0-? resistor is inserted in series, the current drops to 13.8 A. What is the resistance in the original circuit?

Answer 1

Voltage is same in both the circuit ,so

V=I*R

24*R =13.8*(R+25)

24R=13.8R+345

R=33.82 ohms

Answer 2

Let original Resistance be R
we use Ohm's Law, V=IR

in the first case, we have V=24R

since the circuit is in series, the resistances add, so the total resistance in the second case is R+25, so we have

V=13.8(R+25)

assuming the potential difference is the same, we have

24R=13.8(R+25)
24R=13.8R+345
R=33.8

ans: 33.8 ohms

Answer 3

Voltage, current, and resistance follow:

V = IR


Originally, we have:

V = 24 * R

After adding 25 ohms of resistance, we have:

V = 13.8 * (R + 25)

Since V and R are held constant, we have:

24*R = 13.8*(R+25)
24R = 13.8R + 345
10.2R = 345
R = 345/10.2
R = 33.8 OHM