Question

b) A coil of 20 Ω resistance is joined in parallel with a coil of R Ω resistance. This combination is then joined in series with a piece of apparatus A, and the whole circuit connected to 100 V mains. What must be the value of R so that A shall dissipate 600 W with 10 A passing through it?

Answer 1

Answer:

to find the value of R

current given is 10 A

voltage given is 100 V

and power consuming is 600W

A = power = I2R as , 600=10×10 ×R

600÷100=R as , A=6ohm

connection is likely as seen upward.

to find value of R we know that current is same in series case so, total resistance = 20+6+R =26R

as we know that, V=IR

100= 10×26R

10=26R as , R= 2.6 ohm .

so, value of R is 2.6 ohm

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Explanation:

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