A) \[2\Omega \]
B) \[14\Omega \]
C) \[4\Omega \]
D) \[10\Omega \]
Correct Answer: D
Solution :
In series L-R circuit, impedance is given by \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] where R is the resistance and \[{{X}_{L}}\]the inductive reactance. Given, \[R=8\,\Omega ,\,{{X}_{L}}=6\,\Omega \] \[\therefore \] \[Z=\sqrt{{{(8)}^{2}}+{{(6)}^{2}}}\] \[=\sqrt{64+36}\] \[=\sqrt{100}\] \[=10\,\Omega \] Note: Inductive reactance\[{{X}_{L}}\] is zero then circuit is purely resistive.
You need to login to perform this action.
You will be redirected in
3 sec
