A 100v battery is connected to the electric network .If the power consumed in the 2ohm resistor is 200W,determine the power dissipated in the 5ohm resistor?
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you forgot to attach the figure. well, i attached.
the following steps are required to solve this problems :-
step1 :- find equivalent resistance,
R₁ = 30 × 6/(30 + 6) = 5 Ω
R₂ = 10Ω , R' = 5Ω + 5Ω = 10Ω
R₃ = 40Ω , R'' = 10*40/(10 + 40) = 8Ω
R₄ = 2Ω
so, Req = 2Ω + 8Ω = 10Ω
step 2 :- find current through the circuit ,
use Ohm's law, V = iReq
100V = i10Ω ⇒ i = 10 A
step3 :- now, let i₁ current passing through first toop.
now, solve the circuit by using Kirchoff's law,
100 -40i₁ - 2I = 0
⇒ 100 - 40i₁ - 2 × 10 = 0 ⇒ i₁ = 2A
and now, current through 5Ω resistor is (10 - i₁) = 8A
so, power = (current through 5Ω)²(5Ω)
= (8)² × 5 = 320 watt
the following steps are required to solve this problems :-
step1 :- find equivalent resistance,
R₁ = 30 × 6/(30 + 6) = 5 Ω
R₂ = 10Ω , R' = 5Ω + 5Ω = 10Ω
R₃ = 40Ω , R'' = 10*40/(10 + 40) = 8Ω
R₄ = 2Ω
so, Req = 2Ω + 8Ω = 10Ω
step 2 :- find current through the circuit ,
use Ohm's law, V = iReq
100V = i10Ω ⇒ i = 10 A
step3 :- now, let i₁ current passing through first toop.
now, solve the circuit by using Kirchoff's law,
100 -40i₁ - 2I = 0
⇒ 100 - 40i₁ - 2 × 10 = 0 ⇒ i₁ = 2A
and now, current through 5Ω resistor is (10 - i₁) = 8A
so, power = (current through 5Ω)²(5Ω)
= (8)² × 5 = 320 watt
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Explanation:
We know that Power, P =I2 R ⇒ 200 = l2 x 2 l2 = 200/2 = 100 ⇒l= √100 = 10A ∴Current flowing through 2Ω resistor = 10 A Potential drop across 2Ωresistor, V =IR = 10 × 2 = 20 V Equivalent resistance of 30Ω and 6Ω (30 x 6/30+6) = 180/36 = 5Ω ∴Therefore, potential across parallel combination of 40Ω and 10Ω =10 x 8 = 80 V ∴ Current through 5Ωresistor, l = 80/10 = 8A ∴Power dissipated in 5Ω resistor = l2 R = 82 x 5 = 320 WRead more on Sarthaks.com - https://www.sarthaks.com/494858/100-battery-connected-to-the-electric-network-shown-the-power-consumed-the-resistor-is-200
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