Physics, asked by deepaliwalia2872, 1 year ago

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

Answers

Answered by Abhishek9731
4
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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 

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