The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A. What is the resistance in the original circuit?

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R ... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4 (7.4+R) ... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i. e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4 (7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R = 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms