Answer :
To find the equivalent resistance [tex]\( R_{\text{eq}} \)[/tex] when the resistors [tex]\( R_1 \)[/tex], [tex]\( R_2 \)[/tex], and [tex]\( R_3 \)[/tex] are connected in series, we need to sum their individual resistances.
Given the resistances:
- [tex]\( R_1 = 40.0 \, \Omega \)[/tex]
- [tex]\( R_2 = 25.4 \, \Omega \)[/tex]
- [tex]\( R_3 = 70.8 \, \Omega \)[/tex]
In a series circuit, the equivalent resistance [tex]\( R_{\text{eq}} \)[/tex] is the sum of all the resistances in the series. Therefore, we calculate:
[tex]\[ R_{\text{eq}} = R_1 + R_2 + R_3 \][/tex]
Substituting the given values:
[tex]\[ R_{\text{eq}} = 40.0 \, \Omega + 25.4 \, \Omega + 70.8 \, \Omega \][/tex]
Performing the addition:
[tex]\[ R_{\text{eq}} = 40.0 + 25.4 + 70.8 \][/tex]
[tex]\[ R_{\text{eq}} = 136.2 \, \Omega \][/tex]
Therefore, the equivalent resistance is:
[tex]\[ R_{\text{eq}} = 136.2 \, \Omega \][/tex]
Given the resistances:
- [tex]\( R_1 = 40.0 \, \Omega \)[/tex]
- [tex]\( R_2 = 25.4 \, \Omega \)[/tex]
- [tex]\( R_3 = 70.8 \, \Omega \)[/tex]
In a series circuit, the equivalent resistance [tex]\( R_{\text{eq}} \)[/tex] is the sum of all the resistances in the series. Therefore, we calculate:
[tex]\[ R_{\text{eq}} = R_1 + R_2 + R_3 \][/tex]
Substituting the given values:
[tex]\[ R_{\text{eq}} = 40.0 \, \Omega + 25.4 \, \Omega + 70.8 \, \Omega \][/tex]
Performing the addition:
[tex]\[ R_{\text{eq}} = 40.0 + 25.4 + 70.8 \][/tex]
[tex]\[ R_{\text{eq}} = 136.2 \, \Omega \][/tex]
Therefore, the equivalent resistance is:
[tex]\[ R_{\text{eq}} = 136.2 \, \Omega \][/tex]
