A) 11.\[\Omega ,\] 17.5A
B) 30.\[\Omega ,\]6.5A
C) 40.\[\Omega ,\] 5 A
D) 50\[\Omega ,\] 4 A
Correct Answer: D
Solution :
Impedance of L-R circuit is \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}\] \[=\sqrt{{{R}^{2}}+{{(2\pi fL)}^{2}}}\] Given, \[L=\frac{0.4}{\pi }H,\] \[R=30\Omega ,\] \[f=50\,cycle/s\] So, \[Z=\sqrt{{{(30)}^{2}}+{{\left( 2\pi \times 50\times \frac{0.4}{\pi } \right)}^{2}}}\] \[=\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}\] \[=\sqrt{900+1600}\] \[=50\Omega \] and current \[I=\frac{V}{Z}=\frac{200}{50}=4A\]
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