A circuit consisting of three resistances 12, 18 and 36 respectively joined in parallel is connected in series with a fourth resistance. the whole circuit is supplied at 60 v and it is found that power dissipated in 12ohm resistance is 36w.determine the value of fourth resistance and the total power dissipated in the group
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4 answers · Engineering
Best Answer
Always start with what you know
P = E²/R
36 = E²/12
E = 20.78 volts.
this voltage is across all the parallel resistors. Their total resistance is
1/R = 1/12 + 1/18 + 1/36 = 0.08333 + 0.05555 + 0.02777 = 0.16666
R = 6 ohms
with 20.78 volts, the current through all 3 is
I = 20.78/6 = 3.46 amps
that means the series resistor has 60 – 20.78 volts across it or 39.22 volts.
with 3.46 amps, that means it's resistance is 11.34 ohms
total resistance = 11.34 + 6 = 17.34 ohms
total power = 60 x 3.46 = 207.6 watts
thank you if my answer helpful add me as brainleast!
4 answers · Engineering
Best Answer
Always start with what you know
P = E²/R
36 = E²/12
E = 20.78 volts.
this voltage is across all the parallel resistors. Their total resistance is
1/R = 1/12 + 1/18 + 1/36 = 0.08333 + 0.05555 + 0.02777 = 0.16666
R = 6 ohms
with 20.78 volts, the current through all 3 is
I = 20.78/6 = 3.46 amps
that means the series resistor has 60 – 20.78 volts across it or 39.22 volts.
with 3.46 amps, that means it's resistance is 11.34 ohms
total resistance = 11.34 + 6 = 17.34 ohms
total power = 60 x 3.46 = 207.6 watts
thank you if my answer helpful add me as brainleast!
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