Can someone solve these problems? An incandescent light-bulb rated at 40 W will

Question

Can someone solve these problems?

An incandescent light-bulb rated at 40 W will dissipate 40 W as heat and light when connected across a 120-V ideal voltage source. A 60-W bulb will dissipate 60 W when connected across the same source. If the bulbs are connected in series across the same source, determine the power that each of the two bulbs will dissipate. Find the equivalent resistance of the circuit below by combining resistors in series and parallel. Assume R0 = 8 ohm,R1 = 16 ohm, R2 = 12 ohm,R3 = 6 ohm,R4 = 20 ohm, andR5 = 4ohm. Find the equivalent resistance of the circuit below and the current I. Assume Vs = 12 V,R0 = 40 ohm, R1 = 20 ohm, R2 = 80 ohm, R3 = 50 ohm, R4 = 100 ohm, Rs = 120 ohm, and R6 = 60 ohm. Given the following circuit, find the voltage at nodes a and b.

Explanation / Answer

1)

the power loss across a fix voltage is P = V^2/R where V is the voltage and R is the resistenace.

so we can find the resistances of the two bulb.

the equations are

R1 = 120*120/40 = 360 ohm

R2 = 120*120/60 = 240 ohm

so the y are connected in series then net R = R!+R2 = 600 ohm.

so net power loss now P = 120*120/600 = 24 W

2)see R2 and R3 are in series and they are parallel to R5 so we have (R2+R3)||R5 now this is in series with R4 so we have R4 + (R2+R3)||R5. this is in parallel to R1 so R1||(R4 + (R2+R3)||R5) and this is in series with R0

so net Req = R0 + R1||(R4 + (R2+R3)||R5)

3) in asimilar way the equivalent resisteance can be found as

Req = R1 +(R0+R2)||(R3 + (R4||R5||R6))

=39.909 ohm.

after simplification if you use nodal analysis then we have

V/20 + V/26.67 + V/78.57 = 12/20

so V = 5.9866 V at the higher node at where I flows.

I = 0.07483 A

4)

nodal analysis give at a

Va/R1 + (Va-Vb)/R2 +Ia = 0---------------1

and

(Vb-Va)/R2 - Ia-Ib = 0------------------2

solve the two equation and get the result.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---