Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }
Question: 1
29(i).
The total capacitance of the circuit is :
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A quick check for series capacitors: The equivalent capacitance will ALWAYS be mathematically smaller than the smallest individual capacitor in that series chain. Since 2 is less than 3, our answer makes physical sense.
Concept: • When two or more capacitors are connected head-to-tail in a single branch, they are said to be in a series combination.
• The reciprocal of the equivalent (total) capacitance of capacitors in series is equal to the sum of the reciprocals of their individual capacitances.
• For two capacitors in series, the formula simplifies conveniently to the "product over sum" rule: $C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2}$.
Step 1: Identify the capacitor configuration
By carefully examining the provided circuit diagram, we see a specific branch located between node A and node B.
This specific branch exclusively contains the $6 \ \mu$F capacitor and the $3 \ \mu$F capacitor connected end-to-end sequentially.
Therefore, these two capacitors are strictly connected in a series configuration relative to each other.
Step 2: Calculate the total equivalent capacitance
Let $C_1 = 6 \ \mu$F and $C_2 = 3 \ \mu$F.
Apply the series equivalent capacitance formula:
\[ C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} \]
Substitute the given values:
\[ C_{eq} = \frac{6 \times 3}{6 + 3} \]
\[ C_{eq} = \frac{18}{9} \]
\[ C_{eq} = 2 \ \mu\text{F} \] Step 3: Conclusion
The total equivalent capacitance of the capacitor branch in the circuit is exactly $2 \ \mu$F. This directly matches option (D).
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Question: 2
The current in the $10 \ \Omega$ resistor is :
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Whenever a DC circuit problem states "after a long time" or "steady state", simply physically erase or ignore any branches containing capacitors. Analyze what remains purely as a resistor circuit.
Concept: • A fully charged capacitor acts essentially as an open circuit (infinite resistance) to direct current (DC).
• The phrase "connected... for a long time" implies the circuit has completely reached its DC steady state.
• In this steady state, absolutely zero current flows through any branch containing a capacitor.
• Current will exclusively flow through the purely resistive branches, governed by Ohm's Law ($I = \frac{V}{R_{eq}}$).
Step 1: Analyze the steady-state circuit
Because the circuit has been running "for a long time" with a DC battery, the $6 \ \mu$F and $3 \ \mu$F capacitors are fully charged.
Thus, the middle branch between A and B containing these capacitors blocks all DC current. The current in this specific branch is $0$ A.
Therefore, the total steady current from the 3V battery flows out, passes entirely through the top $5 \ \Omega$ resistor, and then must route completely through the $10 \ \Omega$ resistor to return to the battery.
Step 2: Calculate total resistance and total current
Since the current paths through the $5 \ \Omega$ and $10 \ \Omega$ resistors are sequential, they act as a simple series combination.
Total resistance of the active circuit, $R_{eq} = 5 \ \Omega + 10 \ \Omega = 15 \ \Omega$.
The applied battery voltage is $V = 3 \text{ V}$.
Applying Ohm's law to find the main circuit current:
\[ I = \frac{V}{R_{eq}} \]
\[ I = \frac{3 \text{ V}}{15 \ \Omega} \]
\[ I = 0.2 \text{ A} \] Step 3: Conclusion
Because all the active current flows through the $10 \ \Omega$ resistor, the current passing through it is precisely $0.2 \text{ A}$. This matches option (C).
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Question: 3
The potential difference between point A and B is :
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Alternatively, you can view the circuit as a voltage divider. The 3V battery is split across the $5 \ \Omega$ and $10 \ \Omega$ series resistors. Voltage across the $10 \ \Omega$ part is $V_{10} = 3\text{V} \times \frac{10}{10 + 5} = 3 \times \frac{10}{15} = 2\text{V}$. Both methods are equally valid and fast.
Concept: • The potential difference between two specific nodes in a circuit can be found by calculating the voltage drop across any single branch connecting those two nodes.
• Nodes A and B are physically connected by the $10 \ \Omega$ resistor branch.
• According to Ohm's Law, the potential difference across a resistor is exactly the product of the current flowing through it and its resistance ($V = I \times R$).
Step 1: Identify the active component between A and B
From our previous steady-state analysis, we know that nodes A and B have two parallel branches between them.
One branch contains the fully charged capacitors (which blocks DC current).
The other branch contains the $10 \ \Omega$ resistor.
The potential difference $V_{AB}$ must strictly be the voltage drop appearing across this $10 \ \Omega$ resistor.
Step 2: Calculate the voltage drop
We already successfully calculated that the steady current $I$ flowing entirely through the $10 \ \Omega$ resistor is $0.2 \text{ A}$.
Apply Ohm's law specifically across this resistor:
\[ V_{AB} = I_{resistor} \times R \]
\[ V_{AB} = 0.2 \text{ A} \times 10 \ \Omega \]
\[ V_{AB} = 2 \text{ V} \] Step 3: Conclusion
The potential difference heavily maintained between points A and B is exactly $2 \text{ V}$. This definitively matches option (A).
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Question: 4
The value of charge on the plates of the $6 \ \mu$F capacitor is :
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Remember: In series, Charge ($Q$) is the exact same for all components, but Voltage ($V$) divides. In parallel, Voltage ($V$) is the exact same, but Charge ($Q$) divides based on capacity.
Concept: • Capacitors connected in a series configuration inherently store the exact same magnitude of charge ($Q$) on their plates, regardless of their individual capacitance values.
• The total charge stored in this series combination is completely determined by the equivalent capacitance ($C_{eq}$) of the branch and the total potential difference applied directly across that entire branch.
• The relationship is given by the fundamental formula: $Q = C_{eq} \times V_{branch}$.
Step 1: Identify the parameters for the capacitor branch
The capacitor branch is connected directly between node A and node B.
From a previous calculation, we established that the potential difference across nodes A and B is exactly $V_{AB} = 2 \text{ V}$. This is the voltage forcefully applied across the entire series capacitor combination.
We also previously calculated the equivalent capacitance of this series branch to be $C_{eq} = 2 \ \mu$F.
Step 2: Calculate the total charge on the series branch
Use the standard capacitance charge formula:
\[ Q = C_{eq} \times V_{AB} \]
Substitute the known values:
\[ Q = (2 \ \mu\text{F}) \times (2 \text{ V}) \]
\[ Q = 4 \ \mu\text{C} \]
Because the $6 \ \mu$F capacitor and the $3 \ \mu$F capacitor are connected in strict series, they must forcefully share this exact identical charge.
Step 3: Conclusion
The charge residing strictly on the plates of the $6 \ \mu$F capacitor is $4 \ \mu$C. This corresponds directly to option (B).
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Question: 5
The wire between two capacitors is cut at point P. The current in the circuit will :
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This is a classic conceptual trick question. Examiners try to make you think a physical change always implies an electrical change. Always evaluate the pre-change electrical state first!
Concept: • The question asks about the overall current in the resistive portion of the circuit after a physical modification is made.
• It's crucial to evaluate what role the modified branch was actively playing in the circuit just before it was altered.
• In a fully settled DC steady state, a branch containing a capacitor possesses infinite resistance and actively passes completely zero current.
Step 1: Analyze the initial steady state
Before any wires are cut, the circuit has been connected "for a long time," meaning it has achieved a perfect steady state.
In this state, the capacitors are fully charged. As established previously, the entire capacitor branch between node A and node B acts as an open circuit.
The steady-state current flowing through this specific capacitor branch is $0 \text{ A}$.
The only active current in the entire system is the $0.2 \text{ A}$ flowing through the $5 \ \Omega$ and $10 \ \Omega$ resistors.
Step 2: Analyze the effect of cutting the wire
The wire is physically cut at point P, which is located directly between the two series capacitors.
Cutting this wire physically creates an open circuit in that specific branch.
However, electrically, that branch was already acting identically to an open circuit regarding the DC current flow.
Since the branch was previously carrying zero current, physically severing it completely fails to alter the resistance profile or the current distribution of the remaining active resistive loops in the circuit.
Step 3: Conclusion
Because the capacitor branch was entirely inactive in terms of continuous DC current flow, severing it changes nothing. The current in the main circuit will undeniably remain the same at $0.2 \text{ A}$. This makes option (C) correct.
