Question

Consider the circuit with R₁ = 4 Ω, R₂ = 10 Ω, R₃ = 9 Ω, R₄ = 6 Ω, and R₅ = 7 Ω as above with a battery of 10 volt.

(a) Calculate the equivalent resistant of the entire circuit.

(b) Calculate the total current in the circuit.

(c) Calculate the voltage drop across the R₅ resistor.

(d) Calculate the current through the R₅ resistor.

(e) Calculate the dissipated power across the R₅ resistor.

Answer 1

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:

R2 and R3 are in parallel, So

R23 = 10*9/(10 + 9) = 4.74 ohm

R23 and R4 are in series, So

R234 = R23 + R4 = 4.74 + 6 = 10.74 ohm

R234 and R5 are in parallel, So

R2345 = Rp = 10.74*7/(10.74 + 7) = 4.24 ohm

Now Rp and R1 are in series, So

Req = Rp + R1 = 4.24 + 4

Req = 8.24 ohm

Part B.

Using ohm's law:

V = Ieq*Req

Ieq = V/Req = 10/8.24

Ieq = 1.21 Amp

Part 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. Since R1 and Rp are in series, So

I1 = Ip = Ieq

Ip = 1.21 Amp

Vp = Ip*Rp

Vp = 1.21*4.24 = 5.13 V

Now since R234 and R5 are in parallel, So

V5 = V234 = Vp

V5 = Voltage drop across R5 resistor = 5.13 V

Part C.

Using ohm's law:

V5 = I5*R5

I5 = V5/R5 = 5.13/7

I5 = 0.733 Amp = Current in R5 resistor

Part D.

Power dissipated in R5 resistor will be:

P5 = V5*I5

P5 = 5.13*0.733

P5 = 3.76 W = Power dissipated in R5 resistor

Let me know if you've any query.