Question

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Answer 1

✎Question-

Heat energy is being produced in a resistance in a circuit at the rate of 100W. The current of 3A is flowing in the circuit. What must be the value of the resistance?

       

✎To Find-

Resistance-?

     

✎Given-

Heat energy or power = 100W

Current = 3A

           

✎Formulas used -

$\boxed{\sf{P = I^{2}R}}$

or

$\boxed{\sf{P= V \times R}}$

$\boxed{\sf{V= I \times R}}$

where,

P= Power

V= Voltage

R= resistance

I= current

     

✎ SoLution-

First method-

$\sf{P = I^{2}R}$

$\sf{100=3^2 \times R}$

$\sf{\dfrac{100}{9}=R}$

$\sf{R=11.1 \Omega}$

Second method-

Power= V× l

100= V× 3

V=33.3v

Now

By using ohm's law,

V=lR

$\sf{33.3=3 \times R}$

$\sf{R= \dfrac{33.3}{3} }$

$\sf{R=11.1 \Omega}$

$\pink{\underline{\bf{\therefore The\ value\ of\ resistance\ is\ 11 \Omega }}}$

______________________________

Answer 2

Find-

Resistance-?

     

Given-

Heat energy or power = 100W

Current = 3A

           

Formulas used -

\boxed{\sf{P = I^{2}R}}P=I2R

or

\boxed{\sf{P= V \times R}}P=V×R

\boxed{\sf{V= I \times R}}V=I×R

where,

P= Power

V= Voltage

R= resistance

I= current

     

SoLution-

First method-

\sf{P = I^{2}R}P=I2R

\sf{100=3^2 \times R}100=32×R

\sf{\dfrac{100}{9}=R}9100=R

\sf{R=11.1 \Omega}R=11.1Ω

Second method-

Power= V× l

100= V× 3

V=33.3v

Now

By using ohm's law,

V=lR

\sf{33.3=3 \times R}33.3=3×R

\sf{R= \dfrac{33.3}{3} }R=333.3

\sf{R=11.1 \Omega}R=11.1Ω

\pink{\underline{\bf{\therefore The\ value\ of\ resistance\ is\ 11 \Omega }}}∴The value of resistance is 11Ω

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