Question

Two capacitors, one of \( 3\,\mu\text{F} \) and the other of \( 6\,\mu\text{F} \), are connected in series in the circuit for a long time. The total capacitance of the circuit is:

• A. \( 6\,\mu\text{F} \)
• B. \( 3\,\mu\text{F} \)
• C. \( 9\,\mu\text{F} \)
• D. \( 2\,\mu\text{F} \)
• A. \( 0.3\,\text{A} \)
• B. \( 0.6\,\text{A} \)
• C. \( 0.2\,\text{A} \)
• D. \( 0\,\text{A} \)
• A. \( 2\,\text{V} \)
• B. \( 0.3\,\text{V} \)
• C. \( 0.2\,\text{V} \)
• D. \( 3\,\text{V} \)
• A. \( 6\,\mu\text{C} \)
• B. \( 4\,\mu\text{C} \)
• C. \( 12\,\mu\text{C} \)
• D. \( 8\,\mu\text{C} \)
• A. increase
• B. decrease
• C. remain the same
• D. first increase then become stable

Answer 1

The Correct Option is D

Concept: When two or more capacitors are connected together in a series combination, the reciprocal of the equivalent capacitance (\(C_{\text{eq}}\)) of the network is calculated by taking the sum of the reciprocals of the individual capacitances. Mathematically, for two capacitors connected in series, the expression is defined as: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] By taking the common denominator and rearranging the equation, we arrive at the standard product-over-sum formula: \[ C_{\text{eq}} = \frac{C_1 \cdot C_2}{C_1 + C_2} \]

Step 1: State the given individual values of the capacitances from the question text: \[ C_1 = 3\,\mu\text{F} \] \[ C_2 = 6\,\mu\text{F} \]

Step 2: Substitute these numeric values directly into the derived product-over-sum formula: \[ C_{\text{eq}} = \frac{3 \times 6}{3 + 6} \]

Step 3: Calculate the product in the numerator and the sum in the denominator explicitly: \[ \text{Numerator} = 3 \times 6 = 18 \] \[ \text{Denominator} = 3 + 6 = 9 \]

Step 4: Perform the final division to find the total combined equivalent capacitance: \[ C_{\text{eq}} = \frac{18}{9} = 2\,\mu\text{F} \] Hence, the total capacitance of the circuit evaluates precisely to \( 2\,\mu\text{F} \), corresponding to Option (D).

Answer 2

Concept: The problem states that the network has been connected to the DC source for a “long time.” This phrase specifies that the system has reached a steady-state condition. In a steady-state DC circuit:
• A fully charged capacitor offers an infinite capacitive reactance (\(R_C \rightarrow \infty\)) to direct current.
• Therefore, the capacitor behaves as a perfect open circuit, meaning no steady-state conduction current can pass through the branch containing the capacitors.
• The total current leaving the battery flows exclusively through the purely resistive branches of the circuit.

Step 1: Analyze the behavioral state of the capacitor branch. Since the circuit is under steady-state conditions, the current flowing through the branch containing the \( 3\,\mu\text{F} \) and \( 6\,\mu\text{F} \) capacitors drops to exactly zero.

Step 2: Determine the total active resistance of the circuit. The steady-state current originating from the \( 3\,\text{V} \) battery flows continuously in a single closed loop through the \( 5\,\Omega \) resistor and the \( 10\,\Omega \) resistor. Because these two resistors are connected in a simple series loop, their total combined equivalent resistance (\(R_{\text{total}}\)) is calculated via direct summation: \[ R_{\text{total}} = R_1 + R_2 = 5\,\Omega + 10\,\Omega = 15\,\Omega \]

Step 3: Compute the total circuit loop current using Ohm's Law (\(I = \frac{V}{R}\)): \[ I = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{3\,\text{V}}{15\,\Omega} \]

Step 4: Simplify the fraction to find the numerical decimal current value: \[ I = \frac{1}{5}\,\text{A} = 0.2\,\text{A} \] Since the \( 10\,\Omega \) resistor is part of this continuous single main series loop, the current passing straight through it is exactly \( 0.2\,\text{A} \). This matches Option (C).

Answer 3

Concept: In the given parallel circuit configuration, the branch containing the series combination of the two capacitors between nodes A and B is connected directly in parallel across the \( 10\,\Omega \) resistor. According to the basic laws of electrical networks, components connected in parallel share the exact same potential difference across their respective terminals: \[ V_{\text{AB}} = V_{\text{capacitor branch}} = V_{10\,\Omega\text{ resistor}} \]

Step 1: Recall the steady-state loop current that was determined in part (ii): \[ I = 0.2\,\text{A} \]

Step 2: Apply Ohm's Law specifically to the \( 10\,\Omega \) resistor branch to calculate the potential drop across it: \[ V = I \cdot R \] Substitute the loop current and the resistance value: \[ V_{10\,\Omega} = 0.2\,\text{A} \times 10\,\Omega \]

Step 3: Calculate the product to find the voltage drop: \[ V_{10\,\Omega} = 2\,\text{V} \]

Step 4: Conclude the terminal voltage across points A and B. Because the capacitor branch is connected in parallel directly across this resistor, the potential difference between A and B is identical to the voltage drop across the resistor: \[ V_{\text{AB}} = 2\,\text{V} \] Thus, the correct alternative is Option (A).

Answer 4

Concept: When capacitors are connected in a series configuration, the magnitude of the electrostatic charge (\(Q\)) accumulated on the conductive plates of every individual capacitor in that series branch is strictly identical. The charge on each individual capacitor equals the total charge supplied to their equivalent combination: \[ Q_1 = Q_2 = Q_{\text{total}} = C_{\text{eq}} \cdot V_{\text{branch}} \]

Step 1: Gather the previously computed attributes for the capacitor branch across points A and B:
• Equivalent series capacitance of the branch: \( C_{\text{eq}} = 2\,\mu\text{F} \)
• Potential difference maintained across points A and B: \( V_{\text{AB}} = 2\,\text{V} \)

Step 2: Use the primary definition of capacitance to compute the total accumulated charge (\(Q\)) on the series grouping: \[ Q = C_{\text{eq}} \cdot V_{\text{AB}} \]

Step 3: Substitute the corresponding values into the formula: \[ Q = 2\,\mu\text{F} \times 2\,\text{V} \] \[ Q = 4\,\mu\text{C} \]

Step 4: Apply the series charge principle. Since the \( 3\,\mu\text{F} \) and \( 6\,\mu\text{F} \) capacitors are in series, each stores the same charge magnitude. Therefore, the charge on the plates of the \( 6\,\mu\text{F} \) capacitor is precisely \( 4\,\mu\text{C} \), matching Option (B).

Answer 5

Concept: Electric current requires a continuous, uninterrupted closed conduction path to sustain a steady flow of charge carriers. In a DC circuit operating in steady state, a branch containing capacitors blocks DC and has zero current.

Step 1: Analyze the steady-state current conditions prior to cutting the wire. As established in part (ii), the capacitors are fully charged. The current flowing through the capacitor branch containing point P is already: \[ I_{\text{capacitor branch}} = 0\,\text{A} \]

Step 2: Consider the physical effect of cutting the wire at point P. Cutting the connection permanently transforms that branch into a physical open circuit.

Step 3: Evaluate the impact on the rest of the network. Since the current through that branch was already zero, introducing a physical break does not change the impedance or current distribution of the main resistive loop. The primary loop containing the battery, the \( 5\,\Omega \) resistor, and the \( 10\,\Omega \) resistor remains entirely intact and closed.

Step 4: Conclude the final state of the loop current. The main loop current continues to flow unchanged: \[ I = 0.2\,\text{A} \] Thus, the current in the active circuit will remain completely unaltered. This corresponds to Option (C).