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Nurealicious Nurealicious
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Posts: 11
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10 years ago
A circuit consisting of three resistances 12 ?, 18 ?, and 36 ? respectively, joined in parallel is connected in series with a fourth resistance. The supply voltage is 60 V and the power dissipated in the 12 ? resistor is 36 W. Determine the value of the fourth resistance and the total power dissipated in the circuit.
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ff9176ff9176
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Posts: 12
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10 years ago
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wrote...
10 years ago
The givens are that the 12? dissipates 36W therefore 12*36= 432?=20.784 V

So 20.784 across 8? = 2.598A
And 20.784 across 36? = 0.577A
And 20.784 across 12? = 1.732 A
=4.907A total
60V ? 20.784 = 39.216V at R4 ÷ 4.907 = 7.991? maybe 8? is a common value

Hope this properly answers your question


Guru
wrote...
10 years ago
For the 12? resistor, power is 36W and resistance is 12?:
P1 = (V1² / R)
36 = (V1² / 12)
V1² = 36*12 = 432
V1 = 20.78 V

Because they are in parallel, V = 20.78 for the 18? and 36? also, which I will call V2 and V3.  V4 is the fourth resistor's voltage, which is the remainder of the 60 V not used by V1, V2 or V3 (same amount).  I will just use V1.
V4 = Vtotal - V1
60 - 20.78 = 39.22 V

Now to find the current for the 3 known resistors.  I = V / R
I1 = 20.78 / 12 = 1.73 A
I2 = 20.78 / 18 = 1.15 A
I3 = 20.78 / 36 = 0.58 A

Because the three are in series with the fourth, the current for I4 is equal to the three added together:
I4 = I1 + I2 + I3 = 3.46 A

To find the fourth resistance divide its voltage by its current:
R4 = V4 / I4
R4 = 39.22 / 3.46 = 11.34 ?

To find total power take the source voltage times the total current:
Ptotal = Itotal * Vtotal
Ptotal = 3.46A * 60V = 207.6 W

R4 = 11.34?
Ptotal = 207.6W
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