Answer:
Step-by-step explanation:
given,
Power source = 5 V
R₁ = 25 Ω
R₂ = 50 Ω
a) when they are arranged in series
R = R₁ + R₂
R = 25 + 50
R = 75 Ω
b) current = I =
![(V)/(R)](https://img.qammunity.org/2020/formulas/physics/college/yi1c4on3n43s0r319yo69p8qd4c2edqpcj.png)
I =
![(5)/(75)](https://img.qammunity.org/2020/formulas/physics/college/en9qsbk72hp1zna70zprwhivqe2bhiad7k.png)
I = 0.067 A
so, the resistors are connected in series hence current in both the resistor will be 0.067 A.
c) when they are arranged in parallel
![(1)/(R) = (1)/(R_1) + (1)/(R_2)](https://img.qammunity.org/2020/formulas/physics/college/mtb8fkyt2tobbp418fbu8rehq76gs495a4.png)
![(1)/(R) = (1)/(25) + (1)/(50)](https://img.qammunity.org/2020/formulas/physics/college/rx2q4ocqphde8salhb7tkctxl7ks8vpk8y.png)
R = 16.67 Ω
d) current through the battery
![I = (V)/(R)](https://img.qammunity.org/2020/formulas/physics/college/juzvpqodf9abywxtipi5o1mz8sguz73ygn.png)
= 0.3 A
current through 25 Ω resistor =
![(V)/(R) = (5)/(25) = 0.2 A](https://img.qammunity.org/2020/formulas/physics/college/hlew9uxdegte410uqk04lkj9ix77wj39ib.png)
current through 50 Ω resistor =
![(V)/(R) = (5)/(50) = 0.1 A](https://img.qammunity.org/2020/formulas/physics/college/2szagk6lji41r55q0wmlkbvcn73dwluqfw.png)