Answer:
![\boxed{R_(4)=20\Omega}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xe7km40o3agha8xnopodeddelresl2dcbf.png)
Explanation:
If resistors are in series, so the current
is the same in all of them. In this problem we have four resistors. So, we can get a relationship between the Equivalent resistance of series combination and the four resistors as follows:
![R_(eq)=R_(1)+R_(2)+R_(3)+R_(4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pypx030ogzpvz8xsa5vjdz75fbsv6q23ky.png)
is the total resistance
. Moreover:
![R_(1)=5\Omega \\ R_(2)=10\Omega \\ R_(3)=15\Omega](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lzys93ieszr3lplfi73z3ux5rsfnxchhw6.png)
Therefore:
![50=5+10+15+R_(4) \\ \\ R_(4)=50-(5+10+15) \\ \\ R_(4)=50-30 \\ \\ \boxed{R_(4)=20\Omega}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k8nrwq6gas7uznm93ws0xy0zkga5hwjvr5.png)