SSA of RC circuit

Q. For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is $v_i\left(t\right)=\sqrt{20}\sin {10}^{3}t.$ The output voltage $v_0\left(t\right)$ is equal to

1. $sin({10}^{3}t-45°)$
2. $sin({10}^{3}t+45°)$
3. $sin({10}^{3}t-53°)$
4. $sin({10}^{3}t+53°)$

Q. The circuit shown below is driven by a sinusoidal input $v_i=V_{p}cos\left(t/RC\right)$. The steady output $V_0$ is

1. $\left(V_p/3\right)cos\left(t/RC\right)$
2. $\left(V_p/3\right)sin\left(t/RC\right)$
3. $\left(V_p/2\right)cos\left(t/RC\right)$
4. $\left(V_p/2\right)sin\left(t/RC\right)$

Q.

In the circuit shown below, initially the switch $S_1$ is open, the capacitor C1 has a charge of 6 coulomb, and the capacitor C2 has 0 coulomb. After $S_1$ is closed, the charge on C2 in steady state is Coulomb.

1. 4

Q. In the circuit shown, if $v\left(t\right)=2sin\left(1000t\right)$ volts. $R=1k\Omega$ and $C=1\mu F$. then the steady-state current $i\left(t\right)$, in milliamperes (mA), is

1. $3sin\left(1000t\right)+cos\left(1000t\right)$
2. $sin\left(1000t\right)+cos\left(1000t\right)$
3. $sin\left(1000t\right)+3cos\left(1000t\right)$
4. $2sin\left(1000t\right)+2cos\left(1000t\right)$

Q. The steady state output of the circuit shown in the figure is given by
$y\left(t\right)=A\left(\omega \right)sin(\omega t+\varphi (\omega \left)\right)$. If the amplitude $\left|A\right(\omega \left)\right|=0.25$, then the frequency $\omega$ is

1. $\frac{1}{\sqrt{3}RC}$
2. $\frac{2}{\sqrt{3}RC}$
3. $\frac{1}{RC}$
4. $\frac{2}{RC}$

Q. In the circuit shown in the figure, the input signal is $v_i\left(t\right)=5+3cos\omega t$

The steady state output is expressed as $v_0\left(t\right)=P+Qcos(\omega t-\varphi ).$ If $\omega CR=2$, the values of P and Q are

1. $P=0andQ=\frac{6}{\sqrt{5}}$
2. $P=0andQ=\frac{3}{\sqrt{5}}$
3. $P=5andQ=\frac{6}{\sqrt{5}}$
4. $P=5andQ=3$