If in the circuit, power dissipation is 150 W, then R is :

Question

If in the circuit, power dissipation is 150 W, then R is : (A) 2 Ω (B) 6 Ω (C) 5 Ω (D) 4 Ω

Tardigrade Admin · Accepted Answer

P=(V2/R ncl ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0134e389020d5fc6028a77af0836bbcc-.png" /> 150=((15)2/(R ⋅ 2)2+R) ⇒ R=6 Ω

Q. If in the circuit, power dissipation is $150 W$ , then $R$ is :

$2Ω$

$6Ω$

$5Ω$

$4Ω$

$P = \frac{V ^{2}}{R _{n c l}}$

$150 = \frac{( 15 ) ^{2}}{\frac{R \cdot 2}{2 + R}} \Rightarrow R = 6Ω$

Q. If in the circuit, power dissipation is 150W, then R is :

2Ω

6Ω

5Ω

4Ω