Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - True or False (Q.No. 8)
8.
The power dissipated by R4 is 211 mW.
Discussion:
9 comments Page 1 of 1.
Anonymous said:
6 months ago
First, convert R1, R2, R3 delta network into star. We get star resistance as 3/2,3/8,1/2.
Now, by using nodal analysis, find out voltage at R4, it is 20.55, then power is V * V/R = 211.
Now, by using nodal analysis, find out voltage at R4, it is 20.55, then power is V * V/R = 211.
Jeyarr said:
5 years ago
Can you explain how came the total resistance of 1.8k ohms?
Bill said:
8 years ago
Current over R2=185v /4.5k ohm=41.1 mA.
Voltage across R2=4k * 41.1 mA=164.4V.
Voltage across R3 = 185 - 164.4=20.6V.
Power over R3=20.6V*(20.6V/1kohm) = 424.6mW.
Voltage across R2=4k * 41.1 mA=164.4V.
Voltage across R3 = 185 - 164.4=20.6V.
Power over R3=20.6V*(20.6V/1kohm) = 424.6mW.
Nelson said:
9 years ago
By doing with voltage divider:
Voltage output on R3 is 20.56 volts.
Therefore: P = V^2/R4.
= 20.56^2/2K.
= 211 mW.
Voltage output on R3 is 20.56 volts.
Therefore: P = V^2/R4.
= 20.56^2/2K.
= 211 mW.
Saumitra said:
1 decade ago
Solve reversely by balancing circuit using given data. Its so simple their for I can't solved here.
Dusty T said:
1 decade ago
185/1.8k = 0.1019 A.
P = I(3/7.6)^2*R4 because current separates in a parallel circuit.
P = 0.04^2*4000 = 6.47 Watts?
Can somebody explain my mistake?
P = I(3/7.6)^2*R4 because current separates in a parallel circuit.
P = 0.04^2*4000 = 6.47 Watts?
Can somebody explain my mistake?
Dhana said:
1 decade ago
Can you explain how came the total resistance in 1.8K ohm?
Afash said:
1 decade ago
Total Resistance = 1.8k ohm.
Than,
I=185/1.8.
Then,
P = I^2*R.
211 mW.
Than,
I=185/1.8.
Then,
P = I^2*R.
211 mW.
Shankar said:
1 decade ago
Find the total equivalent resistance 'R'.
Find the current in R4.
Power dissipated = I2/R4.
Find the current in R4.
Power dissipated = I2/R4.
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