This problem is a fantastic exercise in understanding the steady-state behavior of basic electrical components—resistors, inductors, and capacitors—under both Direct Current (DC) and Alternating Current (AC) conditions. Let's embark on a journey to decode each circuit and map them to their corresponding voltage-current relationships.
Analyzing the Setup
We are presented with five distinct circuits, labeled (p) through (t). The problem statement gives us a crucial piece of information: the first two circuits are connected to a variable DC voltage source, while the next three are connected to an AC voltage source operating at a frequency of 50 Hz. Our mission is to determine how the voltages V1 and V2 across the components relate to the steady-state or RMS current I.
Circuit (p)
DC Source with Inductor and Capacitor
Imagine a DC circuit containing a capacitor. What happens when you flip the switch? Initially, current flows as the capacitor charges. But once it is fully charged—in the steady state—it acts as an impenetrable wall to direct current. It becomes an open circuit.
Because the capacitor blocks the DC, the steady-state current I is exactly zero. With no current flowing, the voltage drop across the inductor, V1, is zero. Consequently, the entire source voltage V must appear across the open circuit, which is the capacitor. Thus, V2=V. This perfectly aligns with option (C): V1=0,V2=V.
Circuit (q)
DC Source with Inductor and Resistor
Now, let's look at circuit (q), which features an inductor and a resistor connected to a DC source. In a steady-state DC circuit, the current is constant. An ideal inductor only opposes changes in current. Therefore, to a steady DC current, an ideal inductor is nothing more than a simple piece of wire—a short circuit.
Because the inductor acts as a short, the voltage across it, V1, is zero. The current I is non-zero, and the entire source voltage V drops across the resistor. So, V2=IR=V. Since V2 is positive, V2>V1. Also, V2 is directly proportional to I. This makes circuit (q) a match for options (B), (C), and (D).
Circuit (r)
AC Source with Inductor and Resistor
Based on the problem's text, circuit (r) is our first AC circuit. It contains a 6 mH inductor and a 2 Ω resistor. In an AC circuit, an inductor provides a continuous opposition to the alternating current, known as inductive reactance, XL.
Let's calculate this reactance:
Substituting the given values (f=50 Hz and L=6×10−3 H):
XL=2π×50×6×10−3=600π×10−3≈1.88 Ω
The voltage across the inductor is V1=XLI=1.88I. The voltage across the resistor is V2=RI=2I. Clearly, both V1 and V2 are proportional to I. Furthermore, since 2I>1.88I, we have V2>V1. This matches options (A), (B), and (D).
Circuit (s)
AC Source with Inductor and Capacitor
Circuit (s) is an AC circuit with an inductor and a capacitor. We already know the inductive reactance XL is 1.88 Ω, so V1=1.88I. Now, let's find the capacitive reactance, XC, for the 3 μF capacitor:
XC=2π×50×3×10−61=300π×10−61≈1061 Ω
The voltage across the capacitor is V2=XCI=1061I. Both voltages are proportional to I, and V2 is massively greater than V1. This again matches options (A), (B), and (D).
Circuit (t)
AC Source with Resistor and Capacitor
Finally, circuit (t) is an AC circuit with a resistor and a capacitor. Based on the standard interpretation of the problem, the resistor has a value of 1000 Ω. Therefore, V1=RI=1000I. The voltage across the capacitor remains V2=1061I.
Once again, both voltages are proportional to I, and V2 is greater than V1. This matches options (A), (B), and (D).
Conclusion: By systematically applying the principles of steady-state DC and AC reactances, we have successfully mapped every circuit to its corresponding voltage-current characteristics!
