Question

The circuit shown has three identical resistors in series with device for measuring current (an ammeter). The value of the emf shown is 9.99 V,9.99 V, and each resistor has a value of 120 Ω.120 Ω.

Part A: What is the total resistance of this circuit (?eq)?(Req)?

?eq=Req=

ΩΩ

Part B: What is the value of the current (?)(I) flowing out of the emf?

?=I=

A

Part C: What is the value of the voltage drop (Δ?)(ΔV) across the second resistor?

Δ?=ΔV=

Answer 1

Resistance of each resistor, R = 120 Ω

Emf of the battery, V = 9.99 V

Part A

Total resistance when three identical resistors are connected in series,

R_eq = R + R + R = 3 x 120 = 360 Ω

Part B

Current flowing out of the battery, I = V/R = 9.99/360 = 0.028 A

Part C

Voltage drop across the second resistor, ΔV = IR = 0.028 x 120 = 3.36 V