Question

A 5.0 Ω, 10.0 Ω, 12.0 Ω, and 16.0 Ω resistor are connected in series to a 120 V voltage source. What is the current of the circuit?

A) 2.8 A
B) 4.4 A
C) 44 mA
C) 28 mA

Answer 1

2.8 A

Step-by-step explanation:

since the resistances are in series, the net resistance will be,

R = R1 + R2 + R3 + .....

R = ( 5+10+12+16)Ω

R = 43Ω

From Ohm's law,

V = iR

120 = i * 43

i = 120/43 A

i = 2.8A