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.
Prashanth said:
1 decade 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.
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.
Jagat said:
1 decade ago
Power is total of each component's individual power, irrespective of how components are connected. (i.e. Series or parallel). So 6+10=16W is correct answer. Remember: Power is always additive.
Mahanthesh aadi said:
1 decade 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.
ABAR -4744 said:
6 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.
(1)
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
Gk sami said:
1 decade ago
If two series-connected resistors dissipate 6 watts and 10 watts of power, then what is the total power loss? 4 watts
Ravikumar said:
1 decade ago
Total power dissipation in parallel circuit is sum of individual power dissipation so, P=P1+P2 =6+10=16watts.
Chandru.mdl said:
1 decade ago
If two series-connected resistors dissipate 6 watts and 10 watts of power, then what is the total power loss?
Bharat Malik said:
1 decade ago
P=VI and V is constant in parallel circuit. So P is directly propositional to I. So P=P1+P2 So 10+6=16 watts.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers