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Question

Physics Question on Electric Power

The power dissipated in the circuit shown in the figure is 3030 Watts. The value of RR is

A

20Ω20\, \Omega

B

15Ω15\, \Omega

C

10Ω10\, \Omega

D

30Ω30\, \Omega

Answer

10Ω10\, \Omega

Explanation

Solution

P=V2ReqP=\frac{V^{2}}{R_{e q}}
1Req=1R+15=5+R5R\frac{1}{R_{e q}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R}
Req=(5R5+R)P=30WR_{e q}=\left(\frac{5 R}{5+R}\right) P=30\, W
Substituting the values in equation (i)
30=(10)2(5R5+R)30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)}
R=10Ω\Rightarrow R=10\, \Omega