Question

A circuit 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

Answer 1

a) I = 0.0702A = 70.2mA

b) V1 = 3.09V

c) V2 = 1.19V

d) V3 = 7.72V

Step-by-step explanation:

Given that the circuit consist of three series resistors.

For resistors arranged in series, the total resistance R can be given as:

R = R1 + R2 + R3

R1 = 44 ohms

R2 = 17 ohms

R3 = 110 ohms.

R = 44 + 17 + 110 = 171 ohms

V = 12 V

a) The current of a circuit is given by;

Potential difference V = current × total resistance

V = IR

Making I the subject of formula,

I = V/R

I = 12/171 = 0.0702A

I = 7.02mA

b) the potential difference across any resistors is given by:

V = IR

Since the arrangement is parallel, the same current flows through each of the resistors.

V1 = IR1

V1 = 0.0702 × 44

V1 = 3.09V

c) applying the same rule as b above:

V2 = IR2

V2 = 0.0702 × 17

V2 = 1.19V

d) applying the same rule as b above.

V3 = IR3

V3 = 0.0702 × 110

V3 = 7.72V

Answer 2

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Step-by-step 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