Two electric bulbs rated P1 watt V volts and P2 watt V volts are connected in parallel and V volts are applied to it. The total power will be

Question

Two electric bulbs rated P1 watt V volts and P2 watt V volts are connected in parallel and V volts are applied to it. The total power will be (A) (P1+P2) watt (B) √P1 P2 watt (C) (P1 P2/P1+P2) watt (D) (P1+P2/P1 P2) watt

Tardigrade Admin · Accepted Answer

If resistances of bulbs are R1 and R2, respectively, then in parallel (1/RP)=(1/R1)+(1/R2) ⇒ (1/((V2)Pp))=(1 ⋅/((V2)P1))+(1/((V2)P2)) ⇒ PP=P1+P2

Q. Two electric bulbs rated $P_{1}$ watt $V$ volts and $P_{2}$ watt $V$ volts are connected in parallel and $V$ volts are applied to it. The total power will be

$(P_{1} + P_{2})$ watt

$P_{1} P_{2}$ watt

$\frac{P _{1} P _{2}}{P _{1} + P _{2}}$ watt

$\frac{P _{1} + P _{2}}{P _{1} P _{2}}$ watt

Q. Two electric bulbs rated P1​ watt V volts and P2​ watt V volts are connected in parallel and V volts are applied to it. The total power will be

(P1​+P2​) watt

P1​P2​​ watt

P1​+P2​P1​P2​​ watt

P1​P2​P1​+P2​​ watt

If resistances of bulbs are R1​ and R2​, respectively, then in parallel RP​1​=R1​1​+R2​1​ ⇒(Pp​V2​)1​=(P1​V2​)1⋅​+(P2​V2​)1​ ⇒PP​=P1​+P2​

Q. Two electric bulbs rated $P_{1}$ watt $V$ volts and $P_{2}$ watt $V$ volts are connected in parallel and $V$ volts are applied to it. The total power will be

$(P_{1} + P_{2})$ watt

$P_{1} P_{2}$ watt

$\frac{P _{1} P _{2}}{P _{1} + P _{2}}$ watt

$\frac{P _{1} + P _{2}}{P _{1} P _{2}}$ watt