For the circuit shown in the figure, the switch has been open for a very long time. What is the potential drop across the 15 μH inductor just after closing the switch? (b) What is the potential drop across the 70 μF capacitor after the switch has been closed for a very long time?
For the circuit shown in the figure, the switch has been open for a very long time. What is the potential drop across the 15 μH inductor just after closing the switch? (b) What is the potential drop across the 70 μF capacitor after the switch has been closed for a very long time
The Correct Answer and Explanation is:
To accurately determine the potential drops, we must analyze the behavior of the inductor and capacitor in the context of transient and steady-state analysis in DC circuits. Although the actual circuit figure is not visible, we can assume a common configuration: a DC voltage source connected to a series or parallel RLC circuit with a switch, inductor (15 μH), and capacitor (70 μF).
(a) Potential drop across the 15 μH inductor just after closing the switch
Answer: Equal to the full voltage of the source (V).
Explanation: Just after the switch is closed, the inductor resists any change in current due to its property of self-inductance. According to Faraday’s law: VL=LdIdtV_L = L \frac{dI}{dt}
Because the current through an inductor cannot change instantaneously, the rate of change of current (dI/dt)(dI/dt) is initially very large. Thus, immediately after the switch is closed, the inductor acts like an open circuit, preventing current flow. As a result, the entire source voltage appears across the inductor.
For instance, if the source is 10 V, then: VL(t=0+)=10 VV_L(t=0^+) = 10\,\text{V}
This is due to the inductor’s attempt to maintain its initial zero current state.
(b) Potential drop across the 70 μF capacitor after the switch has been closed for a very long time
Answer: Equal to the full voltage of the source (V).
Explanation: After a long time, the capacitor becomes fully charged in a DC circuit, meaning no current flows through the capacitor branch. At this point, it behaves like an open circuit. The voltage across the capacitor equals the supply voltage (assuming it is directly connected across the source through some resistive path).
Hence, after a long time: VC(t→∞)=VV_C(t \to \infty) = V
So, if the source is 10 V: VC=10 VV_C = 10\,\text{V}
Summary:
- (a) Potential drop across 15 μH inductor just after closing switch: V
- (b) Potential drop across 70 μF capacitor after a long time: V
Where V is the source voltage (typically given in the full problem).
