Acircuit consists of a 12.0-v battery connected to three resistors (44ohm , 17ohm and 110ohm ) in series. a. find the current that flows through the battery. i=__mab. find the potential difference across the 44ohm resistor. v1=__vc. find the potential difference across the 17ohm resistor. v2=__vd. find the potential difference across the 110ohm resistor. v3=__v

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Explanation:

Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

From ohm's law

Voltage (V) = current (I) × resistance (R)

I = V/R = 12/171 = 0.0702A = 0.0702×1000mA = 70.2mA

V1 = I×R1 = 0.0702×44 = 3.09V

V2 = I×R2 = 0.0702×17 = 1.19V

V3 = I×R3 = 0.0702×110 = 7.72V

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

i do.

explanation:

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