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Alternating Current

155103 A fully charged capacitor ' $C$ ' with initial charge ' $q_{0}$ ' is connected to a coil of self inductance ' $L$ ' at $t = 0$ . The time at which the energy is stored equally between the electric and the magnetic field is

1

π \sqrt{LC}

2

\frac{π}{4} \sqrt{LC}

3

2 π \sqrt{LC}

4

\sqrt{LC}

Explanation:

B Charge on the capacitor at any time $t$ is
$q = q_{0} \cos ω t$
At time when energy stored equally in electric and magnetic field then at this time Energy of the capacitor $= \frac{1}{2}$ total energy
So, $\frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2} (\frac{1}{2} \frac{q_{0}^{2}}{C})$
$q = \frac{q_{0}}{\sqrt{2}}$
Compare the value of q from equation (i) and (ii),
$\frac{q_{0}}{\sqrt{2}} = q_{0} \cos ω t$
Then, $\cos ω t = \frac{1}{\sqrt{2}}$
$ω t = \frac{π}{4}$
$t = \frac{π}{4 ω}$
$(∵ ω = \frac{1}{\sqrt{LC}})$
So, $t = \frac{π \sqrt{LC}}{4}$

Karnataka CET-2022

Alternating Current

155104 In a series $L C R$ circuit $R = 300 Ω, L = 0.9 H, C$ $= 2.0 μ F$ and $ω = 1000 rad / \sec$ ., then impedance of the circuit is

1

500 Ω

2

400 Ω

3

1300 Ω

4

900 Ω

Explanation:

A Given data, $R = 300 Ω, L = 0.9 H, C = 2 μ F = 2$ $\times 10^{- 6}, ω = 1000 rad / s$
We know that, in series LCR circuits-
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{(300)^{2} + {(1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}})}^{2}}$
$Z = \sqrt{(300)^{2} + {(900 - \frac{1}{2 \times 10^{- 3}})}^{2}}$
$Z = \sqrt{(300)^{2} + (900 - 500)^{2}}$
$Z = \sqrt{90000 + 160000}$
$Z = \sqrt{250000}$
$Z = 500 Ω$

Karnataka CET-2022

Alternating Current

155106 An RLC circuit consists of a $150 Ω$ resistor, 20 $μ F$ capacitor and a $500 mH$ inductor connected in series with a $100 V$ AC supply. The angular frequency of the supply voltage is $400 rad s^{- 1}$ . The phase angle between current and the applied voltage is

1

\tan^{- 1} (0.8)

2

\tan^{- 1} (0.25)

3

\tan^{- 1} (0.6)

4

\tan^{- 1} (0.5)

Explanation:

D Given,
$R = 150 Ω, C = 20 μ F, L = 500 mH$
$V = 100 volt, ω = 400 rad / s$

We know that,
Inductance $(X_{L}) = ω L = 400 \times 500 \times 10^{- 3} = 200 Ω$
And, Capacitance $(X_{C}) = \frac{1}{ω C} = \frac{1}{400 \times 20 \times 10^{- 6}}$
$= \frac{1000}{8} = 125 Ω$
The magnitude of the impedance in the circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Phase angle $(ϕ) = \tan^{- 1} (\frac{X_{L} - X_{C}}{R})$
$ϕ = \tan^{- 1} (\frac{200 - 125}{150}) = \tan^{- 1} (\frac{75}{150})$
$ϕ = \tan^{- 1} (0.5)$

AP EAMCET-05.07.2022

Alternating Current

155110 A resistor of resistance of $100 Ω$ is connected to an $A C$ source $E = 10 \sin (250 π s^{- 1}) t$ .
The energy dissipated as heat during $t = 0$ to $t =$ $1 ms$ is approximately.

1

\frac{0.57}{π} mj

2

\frac{1.141}{π} mj

3

1 mj

4

0.5 mj

Explanation:

A Given that,
Resistance $(R) = 100 Ω$
$E = 10 \sin (250 π s^{- 1}) t$
$E = E_{0} \sin ω t$
On comparison equation (i) and (ii), we get -
$E_{0} = 10 Volt$
Angular frequency $(ω) = 250 π s^{- 1}$
The energy dissipated as heat $(H)$
$H = \frac{E_{rms}^{2}}{R} T$
Energy dissipated of heat during $t = 0$ to $t = 1 ms = 10^{- 3} s$
$H = \int_{0}^{10^{- 3}} \frac{E_{rms}^{2}}{R} d t$
$H = \frac{1}{R} \int_{0}^{10^{- 3}} E_{0}^{2} \sin^{2} ω t dt$
$H = \frac{1}{100} \int_{0}^{10^{- 3}} (10)^{2} \sin^{2} ω tdt$
$H = \frac{100}{100} \int_{0}^{10^{- 3}} \sin^{2} ω tdt$
$H = \int_{0}^{10^{- 3}} [\frac{1 - \cos 2 ω t}{2}] dt$
$H = \frac{1}{2} \int_{0}^{10^{- 3}} dt - \int_{0}^{10^{- 3}} [\frac{\cos 2 ω t}{2}] dt$
$H = \frac{1}{2} [t]_{0}^{10^{- 3}} - \frac{1}{2} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{10^{- 3}}$
$H = \frac{1}{2} [10^{- 3}] - \frac{1}{4} {[\frac{\sin 2 ω t}{ω}]}_{0}^{10^{- 3}}$
$H = \frac{10^{- 3}}{2} - \frac{1}{4} [\frac{\sin (2 \times 250 π \times 10^{- 3})}{250 π}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π} \times \sin \frac{π}{2}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π}]$
$H = \frac{1}{2} \times [\frac{π - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{3.14 - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{1.14}{1000 π}]$
$H = \frac{0.57}{π} mj$

TS EAMCET 19.07.2022

Alternating Current

155103 A fully charged capacitor ' $C$ ' with initial charge ' $q_{0}$ ' is connected to a coil of self inductance ' $L$ ' at $t = 0$ . The time at which the energy is stored equally between the electric and the magnetic field is

1

π \sqrt{LC}

2

\frac{π}{4} \sqrt{LC}

3

2 π \sqrt{LC}

4

\sqrt{LC}

Explanation:

B Charge on the capacitor at any time $t$ is
$q = q_{0} \cos ω t$
At time when energy stored equally in electric and magnetic field then at this time Energy of the capacitor $= \frac{1}{2}$ total energy
So, $\frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2} (\frac{1}{2} \frac{q_{0}^{2}}{C})$
$q = \frac{q_{0}}{\sqrt{2}}$
Compare the value of q from equation (i) and (ii),
$\frac{q_{0}}{\sqrt{2}} = q_{0} \cos ω t$
Then, $\cos ω t = \frac{1}{\sqrt{2}}$
$ω t = \frac{π}{4}$
$t = \frac{π}{4 ω}$
$(∵ ω = \frac{1}{\sqrt{LC}})$
So, $t = \frac{π \sqrt{LC}}{4}$

Karnataka CET-2022

Alternating Current

155104 In a series $L C R$ circuit $R = 300 Ω, L = 0.9 H, C$ $= 2.0 μ F$ and $ω = 1000 rad / \sec$ ., then impedance of the circuit is

1

500 Ω

2

400 Ω

3

1300 Ω

4

900 Ω

Explanation:

A Given data, $R = 300 Ω, L = 0.9 H, C = 2 μ F = 2$ $\times 10^{- 6}, ω = 1000 rad / s$
We know that, in series LCR circuits-
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{(300)^{2} + {(1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}})}^{2}}$
$Z = \sqrt{(300)^{2} + {(900 - \frac{1}{2 \times 10^{- 3}})}^{2}}$
$Z = \sqrt{(300)^{2} + (900 - 500)^{2}}$
$Z = \sqrt{90000 + 160000}$
$Z = \sqrt{250000}$
$Z = 500 Ω$

Karnataka CET-2022

Alternating Current

155106 An RLC circuit consists of a $150 Ω$ resistor, 20 $μ F$ capacitor and a $500 mH$ inductor connected in series with a $100 V$ AC supply. The angular frequency of the supply voltage is $400 rad s^{- 1}$ . The phase angle between current and the applied voltage is

1

\tan^{- 1} (0.8)

2

\tan^{- 1} (0.25)

3

\tan^{- 1} (0.6)

4

\tan^{- 1} (0.5)

Explanation:

D Given,
$R = 150 Ω, C = 20 μ F, L = 500 mH$
$V = 100 volt, ω = 400 rad / s$

We know that,
Inductance $(X_{L}) = ω L = 400 \times 500 \times 10^{- 3} = 200 Ω$
And, Capacitance $(X_{C}) = \frac{1}{ω C} = \frac{1}{400 \times 20 \times 10^{- 6}}$
$= \frac{1000}{8} = 125 Ω$
The magnitude of the impedance in the circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Phase angle $(ϕ) = \tan^{- 1} (\frac{X_{L} - X_{C}}{R})$
$ϕ = \tan^{- 1} (\frac{200 - 125}{150}) = \tan^{- 1} (\frac{75}{150})$
$ϕ = \tan^{- 1} (0.5)$

AP EAMCET-05.07.2022

Alternating Current

155110 A resistor of resistance of $100 Ω$ is connected to an $A C$ source $E = 10 \sin (250 π s^{- 1}) t$ .
The energy dissipated as heat during $t = 0$ to $t =$ $1 ms$ is approximately.

1

\frac{0.57}{π} mj

2

\frac{1.141}{π} mj

3

1 mj

4

0.5 mj

Explanation:

A Given that,
Resistance $(R) = 100 Ω$
$E = 10 \sin (250 π s^{- 1}) t$
$E = E_{0} \sin ω t$
On comparison equation (i) and (ii), we get -
$E_{0} = 10 Volt$
Angular frequency $(ω) = 250 π s^{- 1}$
The energy dissipated as heat $(H)$
$H = \frac{E_{rms}^{2}}{R} T$
Energy dissipated of heat during $t = 0$ to $t = 1 ms = 10^{- 3} s$
$H = \int_{0}^{10^{- 3}} \frac{E_{rms}^{2}}{R} d t$
$H = \frac{1}{R} \int_{0}^{10^{- 3}} E_{0}^{2} \sin^{2} ω t dt$
$H = \frac{1}{100} \int_{0}^{10^{- 3}} (10)^{2} \sin^{2} ω tdt$
$H = \frac{100}{100} \int_{0}^{10^{- 3}} \sin^{2} ω tdt$
$H = \int_{0}^{10^{- 3}} [\frac{1 - \cos 2 ω t}{2}] dt$
$H = \frac{1}{2} \int_{0}^{10^{- 3}} dt - \int_{0}^{10^{- 3}} [\frac{\cos 2 ω t}{2}] dt$
$H = \frac{1}{2} [t]_{0}^{10^{- 3}} - \frac{1}{2} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{10^{- 3}}$
$H = \frac{1}{2} [10^{- 3}] - \frac{1}{4} {[\frac{\sin 2 ω t}{ω}]}_{0}^{10^{- 3}}$
$H = \frac{10^{- 3}}{2} - \frac{1}{4} [\frac{\sin (2 \times 250 π \times 10^{- 3})}{250 π}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π} \times \sin \frac{π}{2}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π}]$
$H = \frac{1}{2} \times [\frac{π - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{3.14 - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{1.14}{1000 π}]$
$H = \frac{0.57}{π} mj$

TS EAMCET 19.07.2022

Alternating Current

155103 A fully charged capacitor ' $C$ ' with initial charge ' $q_{0}$ ' is connected to a coil of self inductance ' $L$ ' at $t = 0$ . The time at which the energy is stored equally between the electric and the magnetic field is

1

π \sqrt{LC}

2

\frac{π}{4} \sqrt{LC}

3

2 π \sqrt{LC}

4

\sqrt{LC}

Explanation:

B Charge on the capacitor at any time $t$ is
$q = q_{0} \cos ω t$
At time when energy stored equally in electric and magnetic field then at this time Energy of the capacitor $= \frac{1}{2}$ total energy
So, $\frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2} (\frac{1}{2} \frac{q_{0}^{2}}{C})$
$q = \frac{q_{0}}{\sqrt{2}}$
Compare the value of q from equation (i) and (ii),
$\frac{q_{0}}{\sqrt{2}} = q_{0} \cos ω t$
Then, $\cos ω t = \frac{1}{\sqrt{2}}$
$ω t = \frac{π}{4}$
$t = \frac{π}{4 ω}$
$(∵ ω = \frac{1}{\sqrt{LC}})$
So, $t = \frac{π \sqrt{LC}}{4}$

Karnataka CET-2022

Alternating Current

155104 In a series $L C R$ circuit $R = 300 Ω, L = 0.9 H, C$ $= 2.0 μ F$ and $ω = 1000 rad / \sec$ ., then impedance of the circuit is

1

500 Ω

2

400 Ω

3

1300 Ω

4

900 Ω

Explanation:

A Given data, $R = 300 Ω, L = 0.9 H, C = 2 μ F = 2$ $\times 10^{- 6}, ω = 1000 rad / s$
We know that, in series LCR circuits-
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{(300)^{2} + {(1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}})}^{2}}$
$Z = \sqrt{(300)^{2} + {(900 - \frac{1}{2 \times 10^{- 3}})}^{2}}$
$Z = \sqrt{(300)^{2} + (900 - 500)^{2}}$
$Z = \sqrt{90000 + 160000}$
$Z = \sqrt{250000}$
$Z = 500 Ω$

Karnataka CET-2022

Alternating Current

155106 An RLC circuit consists of a $150 Ω$ resistor, 20 $μ F$ capacitor and a $500 mH$ inductor connected in series with a $100 V$ AC supply. The angular frequency of the supply voltage is $400 rad s^{- 1}$ . The phase angle between current and the applied voltage is

1

\tan^{- 1} (0.8)

2

\tan^{- 1} (0.25)

3

\tan^{- 1} (0.6)

4

\tan^{- 1} (0.5)

Explanation:

D Given,
$R = 150 Ω, C = 20 μ F, L = 500 mH$
$V = 100 volt, ω = 400 rad / s$

We know that,
Inductance $(X_{L}) = ω L = 400 \times 500 \times 10^{- 3} = 200 Ω$
And, Capacitance $(X_{C}) = \frac{1}{ω C} = \frac{1}{400 \times 20 \times 10^{- 6}}$
$= \frac{1000}{8} = 125 Ω$
The magnitude of the impedance in the circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Phase angle $(ϕ) = \tan^{- 1} (\frac{X_{L} - X_{C}}{R})$
$ϕ = \tan^{- 1} (\frac{200 - 125}{150}) = \tan^{- 1} (\frac{75}{150})$
$ϕ = \tan^{- 1} (0.5)$

AP EAMCET-05.07.2022

Alternating Current

155110 A resistor of resistance of $100 Ω$ is connected to an $A C$ source $E = 10 \sin (250 π s^{- 1}) t$ .
The energy dissipated as heat during $t = 0$ to $t =$ $1 ms$ is approximately.

1

\frac{0.57}{π} mj

2

\frac{1.141}{π} mj

3

1 mj

4

0.5 mj

Explanation:

A Given that,
Resistance $(R) = 100 Ω$
$E = 10 \sin (250 π s^{- 1}) t$
$E = E_{0} \sin ω t$
On comparison equation (i) and (ii), we get -
$E_{0} = 10 Volt$
Angular frequency $(ω) = 250 π s^{- 1}$
The energy dissipated as heat $(H)$
$H = \frac{E_{rms}^{2}}{R} T$
Energy dissipated of heat during $t = 0$ to $t = 1 ms = 10^{- 3} s$
$H = \int_{0}^{10^{- 3}} \frac{E_{rms}^{2}}{R} d t$
$H = \frac{1}{R} \int_{0}^{10^{- 3}} E_{0}^{2} \sin^{2} ω t dt$
$H = \frac{1}{100} \int_{0}^{10^{- 3}} (10)^{2} \sin^{2} ω tdt$
$H = \frac{100}{100} \int_{0}^{10^{- 3}} \sin^{2} ω tdt$
$H = \int_{0}^{10^{- 3}} [\frac{1 - \cos 2 ω t}{2}] dt$
$H = \frac{1}{2} \int_{0}^{10^{- 3}} dt - \int_{0}^{10^{- 3}} [\frac{\cos 2 ω t}{2}] dt$
$H = \frac{1}{2} [t]_{0}^{10^{- 3}} - \frac{1}{2} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{10^{- 3}}$
$H = \frac{1}{2} [10^{- 3}] - \frac{1}{4} {[\frac{\sin 2 ω t}{ω}]}_{0}^{10^{- 3}}$
$H = \frac{10^{- 3}}{2} - \frac{1}{4} [\frac{\sin (2 \times 250 π \times 10^{- 3})}{250 π}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π} \times \sin \frac{π}{2}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π}]$
$H = \frac{1}{2} \times [\frac{π - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{3.14 - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{1.14}{1000 π}]$
$H = \frac{0.57}{π} mj$

TS EAMCET 19.07.2022

Alternating Current

155103 A fully charged capacitor ' $C$ ' with initial charge ' $q_{0}$ ' is connected to a coil of self inductance ' $L$ ' at $t = 0$ . The time at which the energy is stored equally between the electric and the magnetic field is

1

π \sqrt{LC}

2

\frac{π}{4} \sqrt{LC}

3

2 π \sqrt{LC}

4

\sqrt{LC}

Explanation:

B Charge on the capacitor at any time $t$ is
$q = q_{0} \cos ω t$
At time when energy stored equally in electric and magnetic field then at this time Energy of the capacitor $= \frac{1}{2}$ total energy
So, $\frac{1}{2} \frac{q^{2}}{C} = \frac{1}{2} (\frac{1}{2} \frac{q_{0}^{2}}{C})$
$q = \frac{q_{0}}{\sqrt{2}}$
Compare the value of q from equation (i) and (ii),
$\frac{q_{0}}{\sqrt{2}} = q_{0} \cos ω t$
Then, $\cos ω t = \frac{1}{\sqrt{2}}$
$ω t = \frac{π}{4}$
$t = \frac{π}{4 ω}$
$(∵ ω = \frac{1}{\sqrt{LC}})$
So, $t = \frac{π \sqrt{LC}}{4}$

Karnataka CET-2022

Alternating Current

155104 In a series $L C R$ circuit $R = 300 Ω, L = 0.9 H, C$ $= 2.0 μ F$ and $ω = 1000 rad / \sec$ ., then impedance of the circuit is

1

500 Ω

2

400 Ω

3

1300 Ω

4

900 Ω

Explanation:

A Given data, $R = 300 Ω, L = 0.9 H, C = 2 μ F = 2$ $\times 10^{- 6}, ω = 1000 rad / s$
We know that, in series LCR circuits-
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{(300)^{2} + {(1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}})}^{2}}$
$Z = \sqrt{(300)^{2} + {(900 - \frac{1}{2 \times 10^{- 3}})}^{2}}$
$Z = \sqrt{(300)^{2} + (900 - 500)^{2}}$
$Z = \sqrt{90000 + 160000}$
$Z = \sqrt{250000}$
$Z = 500 Ω$

Karnataka CET-2022

Alternating Current

155106 An RLC circuit consists of a $150 Ω$ resistor, 20 $μ F$ capacitor and a $500 mH$ inductor connected in series with a $100 V$ AC supply. The angular frequency of the supply voltage is $400 rad s^{- 1}$ . The phase angle between current and the applied voltage is

1

\tan^{- 1} (0.8)

2

\tan^{- 1} (0.25)

3

\tan^{- 1} (0.6)

4

\tan^{- 1} (0.5)

Explanation:

D Given,
$R = 150 Ω, C = 20 μ F, L = 500 mH$
$V = 100 volt, ω = 400 rad / s$

We know that,
Inductance $(X_{L}) = ω L = 400 \times 500 \times 10^{- 3} = 200 Ω$
And, Capacitance $(X_{C}) = \frac{1}{ω C} = \frac{1}{400 \times 20 \times 10^{- 6}}$
$= \frac{1000}{8} = 125 Ω$
The magnitude of the impedance in the circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Phase angle $(ϕ) = \tan^{- 1} (\frac{X_{L} - X_{C}}{R})$
$ϕ = \tan^{- 1} (\frac{200 - 125}{150}) = \tan^{- 1} (\frac{75}{150})$
$ϕ = \tan^{- 1} (0.5)$

AP EAMCET-05.07.2022

Alternating Current

155110 A resistor of resistance of $100 Ω$ is connected to an $A C$ source $E = 10 \sin (250 π s^{- 1}) t$ .
The energy dissipated as heat during $t = 0$ to $t =$ $1 ms$ is approximately.

1

\frac{0.57}{π} mj

2

\frac{1.141}{π} mj

3

1 mj

4

0.5 mj

Explanation:

A Given that,
Resistance $(R) = 100 Ω$
$E = 10 \sin (250 π s^{- 1}) t$
$E = E_{0} \sin ω t$
On comparison equation (i) and (ii), we get -
$E_{0} = 10 Volt$
Angular frequency $(ω) = 250 π s^{- 1}$
The energy dissipated as heat $(H)$
$H = \frac{E_{rms}^{2}}{R} T$
Energy dissipated of heat during $t = 0$ to $t = 1 ms = 10^{- 3} s$
$H = \int_{0}^{10^{- 3}} \frac{E_{rms}^{2}}{R} d t$
$H = \frac{1}{R} \int_{0}^{10^{- 3}} E_{0}^{2} \sin^{2} ω t dt$
$H = \frac{1}{100} \int_{0}^{10^{- 3}} (10)^{2} \sin^{2} ω tdt$
$H = \frac{100}{100} \int_{0}^{10^{- 3}} \sin^{2} ω tdt$
$H = \int_{0}^{10^{- 3}} [\frac{1 - \cos 2 ω t}{2}] dt$
$H = \frac{1}{2} \int_{0}^{10^{- 3}} dt - \int_{0}^{10^{- 3}} [\frac{\cos 2 ω t}{2}] dt$
$H = \frac{1}{2} [t]_{0}^{10^{- 3}} - \frac{1}{2} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{10^{- 3}}$
$H = \frac{1}{2} [10^{- 3}] - \frac{1}{4} {[\frac{\sin 2 ω t}{ω}]}_{0}^{10^{- 3}}$
$H = \frac{10^{- 3}}{2} - \frac{1}{4} [\frac{\sin (2 \times 250 π \times 10^{- 3})}{250 π}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π} \times \sin \frac{π}{2}]$
$H = \frac{1}{2} [\frac{1}{1000} - \frac{1}{500 π}]$
$H = \frac{1}{2} \times [\frac{π - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{3.14 - 2}{1000 \times π}]$
$H = \frac{1}{2} [\frac{1.14}{1000 π}]$
$H = \frac{0.57}{π} mj$

TS EAMCET 19.07.2022