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 2 of 3.
Ravi said:
1 decade ago
We know that P = V2R relation in parallel circuit v constant. So addition gives total power.
Salman said:
8 years ago
In parallel P=p1+p2.
In series P=p1.p2/p1+p2.
It is same as how we calculate capacitance.
In series P=p1.p2/p1+p2.
It is same as how we calculate capacitance.
Khalid usman pak said:
1 decade ago
In series circuit, always power equals p(total) = p1*p2/(p1+p2). Here the answer is 3.75.
Aman said:
1 decade ago
Power is additive in any configuration of resistive circuit: PTotal = P1 + P2 + . . . Pn
Mahantesh aadi said:
1 decade ago
If in series connected resistors dissipate 6w and 10w, which is right answer?
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
Akshay Prakash said:
1 decade ago
SAMI total power dissipated will be always sum of individual power.
Sujabanu said:
1 decade ago
In parallel circuit, dissipate the power is same. So p1+p2=6+10=16.
Satyendra shgal said:
10 years ago
Power adding in any configuration simply add that Pn = P1 + P2.
Arshad said:
1 decade ago
Can you tell me how it is 4 watts? can you explain with calc?
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