Question

find $i_3(t)$
answer: $i_3(t) = 1.333 \sin t$ a

Explanation:

Step1: Calculate the equivalent resistance of the parallel part

The formula for equivalent resistance \(R_{eq}\) of two resistors \(R_1 = 6\Omega\) and \(R_2=3\Omega\) in parallel is \(\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\).
Substituting the values: \(\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{3}=\frac{1 + 2}{6}=\frac{3}{6}=\frac{1}{2}\), so \(R_{eq}=2\Omega\).

Step2: Calculate the total resistance in the circuit

The total resistance \(R_{total}\) is the sum of the series - connected \(4\Omega\) resistor and the equivalent parallel resistance. So \(R_{total}=4 + 2=6\Omega\).

Step3: Calculate the total current in the circuit

Using Ohm's law \(I=\frac{V}{R}\), where \(V = 12\sin t\) and \(R = 6\). The total current \(i_{total}(t)=\frac{12\sin t}{6}=2\sin t\).

Step4: Calculate the voltage across the parallel part

The voltage across the parallel part \(v_x\) is equal to the voltage across \(R_{eq}\). Using \(v = iR\), with \(i = 2\sin t\) and \(R = 2\), we get \(v_x=2\sin t\times2 = 4\sin t\).

Step5: Calculate \(i_3(t)\)

Using Ohm's law \(i=\frac{v}{R}\) for the \(3\Omega\) resistor. Since \(v = v_x = 4\sin t\) and \(R = 3\), then \(i_3(t)=\frac{4\sin t}{3}\approx1.333\sin t\).

Answer:

\(i_3(t)=\frac{4}{3}\sin t\approx1.333\sin t\space A\)