Question

R1 and R2 are two resistors connected in series to a 12.0 volt battery. If R1 = 100 Ohms what resistance must R2 be so that the current in the circuit is 80.0 mA? 50 Ohms 25 Ohms 17 Ohms 10 Ohm 40 Ohms

Answer 1

50 ohms

Explanation:

Given:

R1 = 100 Ohms

V = 12 V

I = 80 mA = 0.08 A

In a series circuit, the current is always the same and is equal to:

`I=(V)/(R_t)`

The equivalent resistance in a series circuit is the sum of the resistances, therefore:

`\begin{gathered} I=(V)/(R_1+R_2) \\ \text{ We replacing} \\ 0.08=(12)/(100+R_2) \\ 0.08(100+R_2)=12 \\ 8+0.08R_2=12 \\ R_2=(12-8)/(0.08) \\ R_2=50 \end{gathered}`

The resistance must be equal to 50 ohms.