Question

curses 2 25 Out analysis Find the equivalent resistance for this circuit. Write "Ohms" for units. 100 R 0 RS 40 R3 80 Answer

Answer 1

a.)

Remember:

For series combination

Req = R1 + R2 + R3 +...............

for parallel combination

1/Req = 1/R1 + 1/R2 + 1/R3 + ............

for 2 resistors in parallel it will be

Req = R1*R2/(R1+R2)

Using this Information:

Here R4 and R5 are in series, So

R45 = R4 + R5 = 10 + 4 = 14 Ohm

Here R2 and R3 are in series, So

R23 = R2 + R3 = 4 + 8 = 12 Ohm

Now R23 and R45 are in parallel, So

Rp = R23*R45/(R23 + R45)

Rp = 12*14/(14 + 12) = 6.46 Ohm

Now Rp and R1 are in series, So

Req = R1 + Rp

Req = 4.00 + 6.46

Req = 10.46 Ohms

b.)

Using ohm's law:

V = ieq*Req

ieq = Total current = V/Req

ieq = 5/10.46

ieq = 0.478 Amp = current through battery

c.)

Now remember in resistors parallel combination voltage distribution in each part will be same and in series combination current distribution in each resistor will be same. So

ieq = i1 = ip

ip = 0.478 Amp

Vp = ip*Rp = 0.478*6.46 = 3.088 V

Since R23 and R45 are in parallel, So

Vp = V23 = V45 = 3.088 V

V23 = 3.088 V

i23 = V23/R23 = 3.088/12 = 0.257 Amp

Since R2 and R3 are in series, So

i23 = i2 = i3

i2 = 0.257 Amp

V2 = i2*R2 =0.257*4

V2 = 1.028 V

V2 = 1.03 V = Voltage drop across R2 resistor

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