NEET-XII-Physics
39: Alternating Current
- #10A coil has a resistance of 10 Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V,
30πHz. Find the average power consumed in the circuit.Ans : Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, `` f`` = `` \frac{30}{\,\mathrm{\,\pi \,}}\,\mathrm{\,Hz\,}``
Reactance of resistance-inductance circuit `` \left(Z\right)`` is given by,
Z = `` \sqrt{{R}^{2}+{{X}_{L}}^{2}}``
Here, R = resistance of the circuit
XL = Reactance of the pure inductive circuit
Z = `` \sqrt{{R}^{2}+{\left(2\,\mathrm{\,\pi \,}fL\right)}^{2}}``
= `` \sqrt{{\left(10\right)}^{2}+{\left(2\times \,\mathrm{\,\pi \,}\times \frac{30}{\,\mathrm{\,\pi \,}}\times 0.4\right)}^{2}}``
Average power consumed in the circuit (P) is given by,
P = ErmsIrmscos`` \varphi ``
cos`` \varphi `` = `` \frac{R}{Z}``, Irms = `` \frac{{E}_{rms}}{Z}``
`` ``
`` \therefore P=6.5\times \frac{6.5}{Z}\times \frac{R}{Z}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{{\left(\sqrt{{R}^{2}+{\left(\omega L\right)}^{2}}\right)}^{2}}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{100+576}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{676}``
`` \Rightarrow P=0.625=\frac{5}{8}\,\mathrm{\,W\,}``
Page No 330: