The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is :

  • Option 1)

    15 \Omega

  • Option 2)

    10 \Omega

  • Option 3)

    30 \Omega

  • Option 4)

    20 \Omega

 

Answers (1)

As we discussed in

R_{eq}=\frac{5R}{5+R}

Power - P=\frac{V^{2}}{R}\:=> \:30=\frac{(10^{2})}{(\frac{5R}{5+R})}

150R = 100(5+R)         =>  150R = 500+100R

R\:=\:\frac{500}{50}\:=\:10\Omega

 


Option 1)

15 \Omega

This option is incorrect.

Option 2)

10 \Omega

This option is correct.

Option 3)

30 \Omega

This option is incorrect.

Option 4)

20 \Omega

This option is incorrect.

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