Asked by Siboniso
A circuit,consisting of 3 resistances 12,18 & 36 ohms respectively joined in parallel is connected in series with fourth resistance.the whole is supplied at 60 volt & it is found that the power dissipated in the 12Ohm resistance is 36 watt. Determine the value of the fourth resistance & total power dissipated in the group.
Answers
Answered by
Henry
P1 = V^2 / R1,
P1 = V^2 / 12 = 36 W.
V^2 = 12*36 = 432,
V=20.8 Volts. = Voltage across R1,R2,R3
I1 = V1 / R1 = 20.8 / 12 = 1.73A.
I2 = 20.8 / 18 = 1.16A.
I3 = 20.8 / 36 = 0.578A.
I4=I1 + I2 + I3 = 1.73 + 1.16 + 0.578 =
3.47A.
R4=V4 / I4 = (60 - 20.8) / 3.47 = 11.3
Ohms.
Pt=E*It = E*I4 = 60 * 3.47 = 208 Watts.
P1 = V^2 / 12 = 36 W.
V^2 = 12*36 = 432,
V=20.8 Volts. = Voltage across R1,R2,R3
I1 = V1 / R1 = 20.8 / 12 = 1.73A.
I2 = 20.8 / 18 = 1.16A.
I3 = 20.8 / 36 = 0.578A.
I4=I1 + I2 + I3 = 1.73 + 1.16 + 0.578 =
3.47A.
R4=V4 / I4 = (60 - 20.8) / 3.47 = 11.3
Ohms.
Pt=E*It = E*I4 = 60 * 3.47 = 208 Watts.
Answered by
kibet
Draw a circuit diagram showing correct connection of all resistors and supply voltage
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