Question

The variation between V and I with time (t) in two alternative circuits is shown in the above graphs. Which one of the curves of circuit is capacitive and which one is resistive?

Hint:

In a resistor V and I are in phase; in a capacitor the current leads the voltage by \( 90^{\circ} \). Look for which graph shows I peaking before V.

Answer 1

Step 1 (Write the waveforms): Let the applied voltage be \(V = V_0\sin\omega t\). For a resistor, Ohm's law gives \(I = \dfrac{V_0}{R}\sin\omega t\), so \(I\) has the same \(\sin\omega t\) form as \(V\) (phase angle \(0\)). For a capacitor, \(I = C\dfrac{dV}{dt} = \omega C V_0\cos\omega t = \omega C V_0\sin(\omega t + 90^{\circ})\), so \(I\) is shifted ahead of \(V\) by \(90^{\circ}\).
Step 2 (Match forms to the pictures): A zero phase shift means the two sine curves sit on top of each other in timing, which is exactly the left graph. A \(+90^{\circ}\) shift means the current sine starts and peaks earlier than the voltage sine, which is exactly the right graph.
Step 3 (Assign the circuits): Since \(I\) and \(V\) are together on the left, that circuit obeys \(I \propto \sin\omega t\) and is resistive. Since \(I\) leads \(V\) on the right, that circuit obeys \(I \propto \sin(\omega t + 90^{\circ})\) and is capacitive.
\[\boxed{\text{Left} = \text{Resistive},\quad \text{Right} = \text{Capacitive}}\]