Need a problem solved

R0ot · 10-03-2006 02:56PM #1

If the circuit in figure 2 has the following known values; V = 90 volts, R2 = 15 Ω, R3 = 30 Ω and the power dissipated on R1 is 120 watts calculate the value of the current, the resistance R1, the potential drops, the total power dissipated.

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Gurgle · 10-03-2006 03:08PM

Its like this:
> V2 = V3 = V - V1 = I2 x R2 = I3 x R3
> I1 = I2+I3
> I1 * I1 * R1 = 120W
Now plug in the numbers you have for R2 and R3.

dudara · 10-03-2006 03:08PM

Ok, I did this really quickly, so there might be a mistake.

Three equations are obvious

1. 90 = I1*R1 + I3*R3
2. 90 = I1*R1 + I2*R2
3. I1 = I2 + I3

From 1 and 2, it can be seen that I3*R3 = I2*R2 or that I2 = 2*I3. Hence I1 = 3*R3

P1 = V1*I1 = ((I1)^2)*R1. Thus R1 = P1 / (I1^2)

Sub this into 1 and combine with I1 = 3*I3 to get a quadratic.

R0ot · 10-03-2006 06:25PM

Thanks much appreciated

Pherekydes · 12-03-2006 09:39AM

The total power dissipated = 120 + power across the parallel network

This leads to the same quadratic as Dudara's but a little shorter perhaps.

Remember, you can always simplify a circuit by getting the equivalent resistance of parallel sections.

Need a problem solved

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