The current I in the circuit shown is _____.
Applying KVL around the single loop, starting from the 5V source. KCL at the junction puts the current in the second 2k\(\Omega\) resistor at \(I + 1~\mathrm{mA}\):
Consider the circuit shown in the figure. The current I flowing through the 10\(\Omega\) resister is _____.
Let's analyze the right-most loop (Loop-1) containing the 1\(\Omega\) resistor, 2\(\Omega\) resistor, and 3V source. Let the current in this loop be \(I_1\). Apply Kirchhoff's laws in Loop-1:
The current \(I_1 = -1\) A flows only in the right-most loop, independent of the rest of the circuit. Since no current can enter or leave this self-contained loop from the middle branch, the current I must be zero.
For the circuit shown, the locus of the impedance \(Z(j\omega)\) is plotted as \(\omega\) increases from zero to infinity. The values of \(R_1\) and \(R_2\) are:
We analyze the circuit at the two frequency extremes, \(\omega = 0\) and \(\omega = \infty\).
At \(\omega = 0~rad/s\): The capacitor's impedance is \(X_C = \frac{1}{j\omega C} \to \infty\). The capacitor acts as an open circuit. The impedance \(Z(0)\) is simply the two resistors in series: \(Z(0) = R_1 + R_2\). From the graph, at \(\omega \to 0\), \(Z(0) = 5~k\Omega\).
At \(\omega = \infty~rad/s\): The capacitor's impedance is \(X_C = \frac{1}{j\omega C} \to 0\). The capacitor acts as a short circuit, bypassing \(R_2\). The impedance \(Z(\infty)\) is just \(R_1\): \(Z(\infty) = R_1\). From the graph, at \(\omega \to \infty\), \(Z(\infty) = 2~k\Omega\).
Solving: Substitute \(R_1 = 2~k\Omega\) into equation (1):
Consider the circuit shown in the figure with input \(V(t)\) in volts. The sinusoidal steady state current \(I(t)\) flowing through the circuit is shown graphically (where t is in seconds). The circuit element Z can be _____.
From the graphs: \(V(t) = \sin(t) \Rightarrow V_m = 1~V\) and \(\omega = 1~rad/sec\). \(I(t)\) is a sine wave that lags \(V(t)\) and has a peak value \(I_m = \frac{1}{\sqrt{2}}~A\).
Since the current lags the voltage, the element Z must have an inductive reactance. The total impedance of the circuit is \(Z_0 = R + Z = 1\Omega + jX_L\). The magnitude of the total impedance is:
We have \(|Z_0| = \sqrt{R^2 + X_L^2} = \sqrt{1^2 + (\omega L)^2} = \sqrt{2}\). Since \(\omega = 1\):
Squaring both sides:
The element Z is a 1 H inductor.