Question

Q2- in the circuit below you have V = 12 volts, R1 = R4 = 10 R2 =150 R350 using your knowledge of series and parallel circuits and Ohm's law answer the following questions a- what is the total resistance of the circuit? (10 points) b. What is the total current I? (5 points) C- What is the voltage across R1? (5 points) d- What is the voltage across R2? (5 points) I 12 12V R R2 Ry

Answer 1

Answer:-

a) 23.75 Ω

b) 0.505 Ampere

c) 5.05 volts

d) 1.90 volts

Solution:-

as resistances R2 and R3 are in parallel so Thier equivalent resistance will be,

1/R23 = 1/R2 + 1/R3

1/R23 = 1/15 + 1/5

R23 = 15/4 = 3.75 Ω

Now R1, R23 and R4 are in series connection,so their equivalents resistance is,

R = R1 + R23 + R4

R = 10 + 3.75 + 10

R = 23.75 Ω

The total resistance of the circuit is 23.75 Ω.

Using ohm's law,

V = I * R

putting the known values,

I = 12/23.75

I = 0.505 Ampere

Voltage across R1 resistance is given by,

V1 = I * R1

V1 = 0.505 * 10 = 5.05 volts

Similarly for R4 resistance,

V4 = 5.05 volts

the remaining voltage will drop across the parallel combination of R2 and R3 resistances. which is,

V2 = V3 = V - V1 - V4

V2 = 12 - 5.05 - 5.05

V2 = 1.90 volts

The voltage across resistance R2 is 1.90 volts.