A coil of self inductance L=1 H resistance r=4 Ω is connected in series with a variable resistance R and an a.c. source with ω =3 rad/sec as shown in figure. What should be the value of R (in Ω ) so that rate of heat loss through R is maximum?

Question

A coil of self inductance L=1 H resistance r=4 Ω is connected in series with a variable resistance R and an a.c. source with ω =3 rad/sec as shown in figure. What should be the value of R (in Ω ) so that rate of heat loss through R is maximum? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-oyy9aqgtj69vxymu.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

H=i2R=((V0/√(R + r)2 + (ω L) 2))2× R for maximum value of H solve (d H/d R)=0 On solving R=5 Ω

Q. A coil of self inductance $L = 1 H$ & resistance $r = 4 Ω$ is connected in series with a variable resistance $R$ and an a.c. source with $ω = 3 r a d / sec$ as shown in figure. What should be the value of $R$ (in $Ω$ ) so that rate of heat loss through $R$ is maximum?

$H = i^{2} R = (\frac{V _{0}}{( R + r ) ^{2} + ( ω L ) ^{2}})^{2} \times R$ for maximum value of $H$ solve $\frac{d H}{d R} = 0$
On solving $R = 5 Ω$

Q. A coil of self inductance L=1H & resistance r=4Ω is connected in series with a variable resistance R and an a.c. source with ω=3rad/sec as shown in figure. What should be the value of R (in Ω ) so that rate of heat loss through R is maximum?

H=i2R=((R+r)2+(ωL)2​V0​​)2×R for maximum value of H solve dRdH​=0 On solving R=5Ω

Q. A coil of self inductance $L = 1 H$ & resistance $r = 4 Ω$ is connected in series with a variable resistance $R$ and an a.c. source with $ω = 3 r a d / sec$ as shown in figure. What should be the value of $R$ (in $Ω$ ) so that rate of heat loss through $R$ is maximum?

$H = i^{2} R = (\frac{V _{0}}{( R + r ) ^{2} + ( ω L ) ^{2}})^{2} \times R$ for maximum value of $H$ solve $\frac{d H}{d R} = 0$
On solving $R = 5 Ω$