Question 69: AC fundamentals

Question

In the given circuit, the AC source has $$\omega$$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

Accepted Answer

Here, $$\Omega$$ = 100 rad/s, L = 0.5 H, C = 100 $$\mu$$F, V = 20 V
$$	herefore$$ $${X_L} = \omega L = 100 	imes 0.5 = 50\,\Omega $$
$${X_C} = {1 \over {\omega C}} = {1 \over {100 	imes 100 	imes {{10}^{ - 6}}}} = 100\,\Omega $$
Impedance across capacitor,
$${Z_1} = \sqrt {{R^2} + X_C^2} $$
$$ = \sqrt {{{(100)}^2} + {{(100)}^2}} $$
$${Z_1} = 100\sqrt 2 \,\Omega $$
$$	herefore$$ $${I_1} = {{20} \over {100\sqrt 2 }} = {1 \over {5\sqrt 2 }}\,A$$
Voltage across 100 $$\Omega$$
$$V = {I_1} 	imes 100 = {1 \over {5\sqrt 2 }} 	imes 100 = 10\sqrt 2 \,V$$
Impedance across inductance,
$${Z_2} = \sqrt {{R^2} + {{({X_L})}^2}}  = \sqrt {{{(50)}^2} + {{(50)}^2}} $$
$${Z_2} = 50\sqrt 2 \,\Omega $$
$$	herefore$$ $${I_2} = {{20} \over {50\sqrt 2 }} = {2 \over {5\sqrt 2 }} = {{\sqrt 2 } \over 5}$$
Now, voltage across $$50\,\Omega  = {{\sqrt 2 } \over 5} 	imes 50 = 10\sqrt 2 $$
$${I_1} = {1 \over {5\sqrt 2 }}A$$ at 45$$^\circ$$ leading
$${I_2} = {{\sqrt 2 } \over 5}A$$ at 45$$^\circ$$ lagging
$$	herefore$$ Current through circuit
$${I_{net}} = \sqrt {I_1^2 + I_2^2}  = \sqrt {{{\left( {{1 \over {5\sqrt 2 }}} ight)}^2} + \left( {{{\sqrt 2 } \over 5}} ight)}  = 0.3\,A$$

Answer

The current through the circuit, I is 0.3 A.

Answer

The current through the circuit, I is 0.3$$\sqrt2$$ A.

Answer

The voltage across 100 $$\Omega$$ resistor = 10$$\sqrt2$$ V.

Answer

The voltage across 50 $$\Omega$$ resistor = 10 V.

AC fundamentals

In the given circuit, the AC source has $\omega$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

Choose the correct answer: