A 52-V battery powers a pair of 36Ω resistor connected in series.What is the current flowing through the system? round to the nearest tenth

Question

1.A 52-V battery powers a pair of 36Ω resistor connected in series.What is the current flowing through the system? round to the nearest tenth.

2.A 52-V battery powers a pair of 36Ω resistor connected in parallel .What is the current flowing through the system?

3.Compare the answers to question 1 and 2. Discuss similarities and differences

Answer 1

(1)

Divide the voltage by the total resistance for 2 resistors in series. That gives the current.

I = V/(R1 + R2) = 52/(36 + 36) amps = ? amps

You can finish from here.

(2)

Divide the voltage by the total resistance for 2 resistors in parallel. That gives the current.

1/R = 1/R1 + 1/R2 = (1/36 + 1/36) ohms^-1 = 1/18 ohms^-1

I = V/R = 52(1/18) amps = ? amps

You can finish from here.

(3)

This will be self-evident once you have the results from parts (1) and (2).