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.
ABAR -4744 said:
3 years ago
The circuit may be whatever(series or parallel) But, always total power losses calculated by the addition of power loaded across every element in circuit.
Salman said:
5 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.
Mohan said:
6 years ago
If it series what is the answer?
Satyendra shgal said:
7 years ago
Power adding in any configuration simply add that Pn = P1 + P2.
Fissh likelike said:
8 years ago
Power lose is always additive what ever series or parallel.
Prashanth said:
8 years ago
Total Power in Series and Parallel resistive circuit will be addition of individual powers:
Parallel: P = V^2/(R1||R2) = V^2/((R1*R2)/(R1+R2)).
= V^2/R1+V^2/R2 = P1+P2.
Series: P = I^2*(R1+R2) = I^2*R1+I^2*R2 = P1+P2.
Parallel: P = V^2/(R1||R2) = V^2/((R1*R2)/(R1+R2)).
= V^2/R1+V^2/R2 = P1+P2.
Series: P = I^2*(R1+R2) = I^2*R1+I^2*R2 = P1+P2.
Mahantesh aadi said:
8 years ago
If in series connected resistors dissipate 6w and 10w, which is right answer?
Mahanthesh aadi said:
8 years ago
Which answer is right? Somebody has telling 4w is right, somebody has telling 3.75w is right, somebody has telling 16w is right, please tell me which one is right.
Ravi said:
8 years ago
We know that P = V2R relation in parallel circuit v constant. So addition gives total power.
Khalid usman pak said:
8 years ago
In series circuit, always power equals p(total) = p1*p2/(p1+p2). Here the answer is 3.75.
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