To find the current flowing out of the battery when a 20.0 Ohm resistor and a 60.0 Ohm resistor are connected in series to a 9.00 V battery, we can follow these steps:
1. Identify the resistances and the voltage:
- The first resistor [tex]\( R1 \)[/tex] has a resistance of 20.0 Ohms.
- The second resistor [tex]\( R2 \)[/tex] has a resistance of 60.0 Ohms.
- The voltage [tex]\( V \)[/tex] of the battery is 9.00 V.
2. Calculate the total resistance in the series circuit:
- For resistors in series, the total resistance [tex]\( R_{total} \)[/tex] is the sum of the individual resistances.
[tex]\[
R_{total} = R1 + R2
\][/tex]
Substituting the values:
[tex]\[
R_{total} = 20.0 \, \text{Ohms} + 60.0 \, \text{Ohms}
\][/tex]
[tex]\[
R_{total} = 80.0 \, \text{Ohms}
\][/tex]
3. Use Ohm's Law to calculate the current:
- Ohm's Law states that [tex]\( I = \frac{V}{R} \)[/tex], where [tex]\( I \)[/tex] is the current, [tex]\( V \)[/tex] is the voltage, and [tex]\( R \)[/tex] is the resistance.
[tex]\[
I = \frac{V}{R_{total}}
\][/tex]
Substituting the values:
[tex]\[
I = \frac{9.00 \, \text{V}}{80.0 \, \text{Ohms}}
\][/tex]
4. Calculate the current:
[tex]\[
I = 0.1125 \, \text{A}
\][/tex]
Therefore, the current flowing out of the battery is [tex]\( 0.1125 \)[/tex] A.
