A sinusoidal A.C. current flows through a resistor of resistance 10 Ω. If the peak current is 2 A flowing through the resistor then the power dissipated in W.

Question

A sinusoidal A.C. current flows through a resistor of resistance 10 Ω. If the peak current is 2 A flowing through the resistor then the power dissipated in W. (A) 30 (B) 20 (C) 10 (D) 40

Tardigrade Admin · Accepted Answer

Resistance R=10 Ω Peak value of current I0=2 A Rms value of current I rms =(I0/√2)=(2/√2)=√2 A Power dissipated P=Ir m s2 R ⇒ P=(√2)2 × 10=20 W

Q. A sinusoidal A.C. current flows through a resistor of resistance $10 Ω$ . If the peak current is $2 A$ flowing through the resistor then the power dissipated in _______ $W$ .

30

20

10

40

Resistance $R = 10 Ω$
Peak value of current $I_{0} = 2 A$
Rms value of current $I_{r m s} = \frac{I _{0}}{2} = \frac{2}{2} = 2 A$
Power dissipated $P = I_{r m s}^{2} R$
$\Rightarrow P = (2)^{2} \times 10 = 20 W$

Q. A sinusoidal A.C. current flows through a resistor of resistance 10Ω. If the peak current is 2A flowing through the resistor then the power dissipated in _______W.

30

20

10

40

Resistance R=10Ω Peak value of current I0​=2A Rms value of current Irms​=2​I0​​=2​2​=2​A Power dissipated P=Irms2​R ⇒P=(2​)2×10=20W

Q. A sinusoidal A.C. current flows through a resistor of resistance $10 Ω$ . If the peak current is $2 A$ flowing through the resistor then the power dissipated in _______ $W$ .

Resistance $R = 10 Ω$
Peak value of current $I_{0} = 2 A$
Rms value of current $I_{r m s} = \frac{I _{0}}{2} = \frac{2}{2} = 2 A$
Power dissipated $P = I_{r m s}^{2} R$
$\Rightarrow P = (2)^{2} \times 10 = 20 W$