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Q. A lamp in which $ 10\,\,amp $ current can flow at $ 15\,\,V $ , is connected with an alternating source of potential $ 220\,\,V $ . The frequency of source is $ 50\,Hz $ . The impedance of choke coil required to light the bulb is:

Punjab PMETPunjab PMET 2003Alternating Current

Solution:

The resistance of lamp
$R=\frac{V}{I}=\frac{15}{10}=1.5 \,\Omega$
(Given : $V=15$ volt, $I=10 amp$, A.C. voltage $=220 V , f=50 Hz$ ) The impedance of circuit
$Z=\frac{V}{I}=\frac{220}{10}=22\,ohm$
Again impedance $Z=\sqrt{R^{2}+\omega^{2} L^{2}}$
or $22=\sqrt{(1.5)^{2}+(2 \times \pi \times 50)^{2} L^{2}}$
or $(22)^{2}=(1.5)^{2}+4 \pi^{2} \times 50^{2} \times L^{2}$
or $L=0.07$ henry