Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 9)
9.
If two parallel-connected resistors dissipate 6 watts and 10 watts of power, then what is the total power loss?
Discussion:
28 comments Page 1 of 3.
Prakasam said:
1 decade ago
Power= P1+P2=6+10=16
Shiv said:
1 decade ago
P1=(V*V)/R1=6
P2=(V*V)/R2=10
=>R1=(5/3)R2
R=R1||R2=(5/8)R2
P=(V*V)/R=16
P2=(V*V)/R2=10
=>R1=(5/3)R2
R=R1||R2=(5/8)R2
P=(V*V)/R=16
Sujabanu said:
1 decade ago
In parallel circuit, dissipate the power is same. So p1+p2=6+10=16.
Sandeep said:
1 decade ago
@Sujabanu
If power in series ckt's then o/p power. ?
If power in series ckt's then o/p power. ?
Abbu said:
1 decade ago
Then take current constant and calculate resistance.
Aman said:
1 decade ago
Power is additive in any configuration of resistive circuit: PTotal = P1 + P2 + . . . Pn
Chandrasekhar said:
1 decade ago
Total power loss in a circuit is nothing but the power loss in the individual components =component1 power + component2 power loss.
The above statement is said because the energy cannot be created or destroyed.
The above statement is said because the energy cannot be created or destroyed.
Sriram said:
1 decade ago
I go with Chandrasekhar.
Nisha said:
1 decade ago
TOTAL POWER P=V^2/(R1!!R2)
=v^2*(R1+R2/R1*R2)
=V^2/R2+V^2/R1
=P1+P2
=16W
=v^2*(R1+R2/R1*R2)
=V^2/R2+V^2/R1
=P1+P2
=16W
Ravikumar said:
1 decade ago
Total power dissipation in parallel circuit is sum of individual power dissipation so, P=P1+P2 =6+10=16watts.
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