201. Study the table for three capacitors connected in parallel across the same source.
| Capacitor | Capacitance | Voltage across it | Charge stored |
|---|
| P | \(2C\) | \(V\) | \(2CV\) |
| Q | \(3C\) | \(V\) | \(3CV\) |
| R | \(5C\) | \(V\) | \(5CV\) |
What is the equivalent capacitance?
ⓐ. \(\frac{C}{10}\)
ⓑ. \(30C\)
ⓒ. \(10C\)
ⓓ. \(\frac{10C}{3}\)
Correct Answer: \(10C\)
Explanation: In a parallel combination, the voltage across every capacitor is the same. The total charge is the sum of the individual charges. From the table, \(Q_P=2CV\), \(Q_Q=3CV\), and \(Q_R=5CV\). Therefore, \(Q_{\text{total}}=2CV+3CV+5CV=10CV\). The equivalent capacitance is defined by \(Q_{\text{total}}=C_{\text{eq}}V\). Comparing \(10CV=C_{\text{eq}}V\), we get \(C_{\text{eq}}=10C\). The result is the same as directly adding \(2C+3C+5C\).
202. A graph is plotted for capacitors connected in parallel across the same voltage \(V\). The vertical axis is charge \(Q\) on each capacitor and the horizontal axis is capacitance \(C\). What does the slope of the \(Q\)-versus-\(C\) graph represent?
ⓐ. The reciprocal voltage \(\frac{1}{V}\)
ⓑ. The common voltage \(V\)
ⓒ. The total energy stored
ⓓ. The equivalent capacitance \(C_{\text{eq}}\)
Correct Answer: The common voltage \(V\)
Explanation: \( \textbf{Charge relation:} \)
\[
Q=CV
\]
\( \textbf{Graph axes:} \) The vertical axis is \(Q\), and the horizontal axis is \(C\).
\( \textbf{Constant quantity:} \) In this comparison, all capacitors are connected across the same voltage \(V\).
\( \textbf{Straight-line form:} \)
\[
Q=(V)C
\]
\( \textbf{Slope comparison:} \) Comparing with \(y=mx\), \(Q\) is \(y\), \(C\) is \(x\), and \(V\) is the slope.
\( \textbf{Physical meaning:} \) The graph shows that at fixed voltage, charge stored is directly proportional to capacitance.
\( \textbf{Final answer:} \) The slope represents the common voltage \(V\).
203. Which row correctly compares capacitor combinations with resistor combinations?
| Row | Capacitors in parallel | Resistors in parallel |
|---|
| P | \(C_{\text{eq}}=C_1+C_2+\cdots\) | \(\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots\) |
| Q | \(\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots\) | \(R_{\text{eq}}=R_1+R_2+\cdots\) |
| R | \(C_{\text{eq}}\) is less than smallest \(C\) | \(R_{\text{eq}}\) is greater than largest \(R\) |
| S | Same charge on each | Same current through each |
ⓐ. Row S
ⓑ. Row P
ⓒ. Row R
ⓓ. Row Q
Correct Answer: Row P
Explanation: Capacitors in parallel add directly: \(C_{\text{eq}}=C_1+C_2+\cdots\). Resistors in parallel combine by reciprocal addition: \(\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots\). Row Q gives the series rule for capacitors and the series rule for resistors, not the parallel comparison. Row R is wrong because capacitors in parallel have equivalent capacitance greater than each individual capacitance, while resistors in parallel have equivalent resistance smaller than the smallest resistance. Row S is also wrong because capacitors in parallel have the same voltage, not necessarily the same charge. A common misconception is to assume capacitors follow the same combination pattern as resistors.
204. Three capacitors \(C\), \(2C\), and \(3C\) are connected in parallel to a battery of voltage \(V\). What fraction of the total charge is stored on the capacitor \(3C\)?
ⓐ. \(\frac{1}{2}\)
ⓑ. \(\frac{1}{3}\)
ⓒ. \(\frac{3}{5}\)
ⓓ. \(\frac{2}{3}\)
Correct Answer: \(\frac{1}{2}\)
Explanation: \( \textbf{Parallel condition:} \) All three capacitors have the same voltage \(V\).
\( \textbf{Charge on \(C\):} \)
\[
Q_1=CV
\]
\( \textbf{Charge on \(2C\):} \)
\[
Q_2=2CV
\]
\( \textbf{Charge on \(3C\):} \)
\[
Q_3=3CV
\]
\( \textbf{Total charge:} \)
\[
Q_{\text{total}}=CV+2CV+3CV=6CV
\]
\( \textbf{Required fraction:} \)
\[
\frac{Q_3}{Q_{\text{total}}}=\frac{3CV}{6CV}
\]
\( \textbf{Simplification:} \)
\[
\frac{Q_3}{Q_{\text{total}}}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The capacitor \(3C\) stores \(\frac{1}{2}\) of the total charge.
205. Two capacitors are connected in parallel. Capacitor \(C_1\) stores \(30\,\mu\text{C}\) and capacitor \(C_2\) stores \(45\,\mu\text{C}\) when connected to the same \(15\,\text{V}\) source. What are \(C_1\), \(C_2\), and \(C_{\text{eq}}\)?
ⓐ. \(3.0\,\mu\text{F}\), \(2.0\,\mu\text{F}\), \(5.0\,\mu\text{F}\)
ⓑ. \(2.0\,\mu\text{F}\), \(3.0\,\mu\text{F}\), \(1.2\,\mu\text{F}\)
ⓒ. \(30\,\mu\text{F}\), \(45\,\mu\text{F}\), \(75\,\mu\text{F}\)
ⓓ. \(2.0\,\mu\text{F}\), \(3.0\,\mu\text{F}\), \(5.0\,\mu\text{F}\)
Correct Answer: \(2.0\,\mu\text{F}\), \(3.0\,\mu\text{F}\), \(5.0\,\mu\text{F}\)
Explanation: \( \textbf{Given:} \) \(Q_1=30\,\mu\text{C}\), \(Q_2=45\,\mu\text{C}\), and common voltage \(V=15\,\text{V}\).
\( \textbf{Parallel condition:} \) Both capacitors have the same voltage across them.
\( \textbf{Capacitance relation:} \)
\[
C=\frac{Q}{V}
\]
\( \textbf{Capacitance of \(C_1\):} \)
\[
C_1=\frac{30\,\mu\text{C}}{15\,\text{V}}=2.0\,\mu\text{F}
\]
\( \textbf{Capacitance of \(C_2\):} \)
\[
C_2=\frac{45\,\mu\text{C}}{15\,\text{V}}=3.0\,\mu\text{F}
\]
\( \textbf{Equivalent capacitance in parallel:} \)
\[
C_{\text{eq}}=C_1+C_2
\]
\( \textbf{Final calculation:} \)
\[
C_{\text{eq}}=2.0\,\mu\text{F}+3.0\,\mu\text{F}=5.0\,\mu\text{F}
\]
\( \textbf{Final answer:} \) \(C_1=2.0\,\mu\text{F}\), \(C_2=3.0\,\mu\text{F}\), and \(C_{\text{eq}}=5.0\,\mu\text{F}\).
206. Two capacitors \(C_1=2C\) and \(C_2=5C\) are connected in parallel across the same battery of voltage \(V\). What is the ratio of the energies stored in them?
ⓐ. \(U_1:U_2=5:2\)
ⓑ. \(U_1:U_2=2:5\)
ⓒ. \(U_1:U_2=1:1\)
ⓓ. \(U_1:U_2=4:25\)
Correct Answer: \(U_1:U_2=2:5\)
Explanation: \( \textbf{Parallel condition:} \) Both capacitors have the same potential difference \(V\).
\( \textbf{Energy formula at fixed voltage:} \)
\[
U=\frac{1}{2}CV^2
\]
\( \textbf{Energy in \(C_1\):} \)
\[
U_1=\frac{1}{2}(2C)V^2=CV^2
\]
\( \textbf{Energy in \(C_2\):} \)
\[
U_2=\frac{1}{2}(5C)V^2=\frac{5}{2}CV^2
\]
\( \textbf{Ratio:} \)
\[
U_1:U_2=CV^2:\frac{5}{2}CV^2
\]
\( \textbf{Simplification:} \)
\[
U_1:U_2=2:5
\]
\( \textbf{Physical meaning:} \) In parallel, larger capacitance stores more energy because the voltage is the same.
\( \textbf{Final answer:} \) \(U_1:U_2=2:5\).
207. Three capacitors \(C_1\), \(C_2\), and \(C_3\) are connected in parallel across a battery of voltage \(V\). Which expression gives the total energy stored?
ⓐ. \(\frac{1}{2}\frac{V^2}{C_1+C_2+C_3}\)
ⓑ. \(\frac{1}{2}\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)V^2\)
ⓒ. \(\frac{1}{2}(C_1C_2C_3)V^2\)
ⓓ. \(\frac{1}{2}(C_1+C_2+C_3)V^2\)
Correct Answer: \(\frac{1}{2}(C_1+C_2+C_3)V^2\)
Explanation: In a parallel combination, each capacitor has the same potential difference \(V\). The energy stored in each capacitor is \(U_i=\frac{1}{2}C_iV^2\). Therefore, the total energy is the sum of individual energies. This gives \(U_{\text{total}}=\frac{1}{2}C_1V^2+\frac{1}{2}C_2V^2+\frac{1}{2}C_3V^2\). Taking \(\frac{1}{2}V^2\) common gives \(U_{\text{total}}=\frac{1}{2}(C_1+C_2+C_3)V^2\). Since \(C_{\text{eq}}=C_1+C_2+C_3\) in parallel, the result is also \(U_{\text{total}}=\frac{1}{2}C_{\text{eq}}V^2\). The reciprocal capacitance expression belongs to series combination, not parallel.
208. In a series combination of capacitors, which quantity is the same on each capacitor?
ⓐ. Capacitance of each capacitor
ⓑ. Energy stored in each capacitor
ⓒ. Charge magnitude on each capacitor
ⓓ. Potential difference across each capacitor
Correct Answer: Charge magnitude on each capacitor
Explanation: In a series combination, capacitors are connected one after another so that charge redistribution makes the same charge magnitude appear on each capacitor. If a charge \(+Q\) appears on one plate of a capacitor, induction produces corresponding charges on adjacent plates. Thus each capacitor in series carries the same charge magnitude \(Q\). The potential differences across the capacitors are generally different because \(V=\frac{Q}{C}\). Smaller capacitance gets larger potential difference for the same charge. A common mistake is to apply the parallel rule of same voltage to a series combination.
209. Which expression gives the equivalent capacitance of capacitors \(C_1\), \(C_2\), and \(C_3\) connected in series?
ⓐ. \(1/C_{\text{eq}}=1/C_1+1/C_2+1/C_3\)
ⓑ. \(1/C_{\text{eq}}=C_1+C_2+C_3\)
ⓒ. \(C_{\text{eq}}=C_1+C_2+C_3\)
ⓓ. \(C_{\text{eq}}=1/(C_1^{-1}+C_2+C_3)\)
Correct Answer: \(1/C_{\text{eq}}=1/C_1+1/C_2+1/C_3\)
Explanation: In a series combination, each capacitor carries the same charge \(Q\). The total potential difference is the sum of potential differences across individual capacitors. Since \(V_i=\frac{Q}{C_i}\), the total voltage is \(V=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\). For the equivalent capacitor, \(V=\frac{Q}{C_{\text{eq}}}\). Equating these expressions and cancelling \(Q\) gives \(\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\). The direct-addition formula belongs to parallel capacitors, not series capacitors. The series rule follows from voltage addition and same charge.
210. Two capacitors \(3.0\,\mu\text{F}\) and \(6.0\,\mu\text{F}\) are connected in series across a \(12\,\text{V}\) battery. What are the equivalent capacitance and charge on each capacitor?
ⓐ. \(4.5\,\mu\text{F}\), \(54\,\mu\text{C}\)
ⓑ. \(2.0\,\mu\text{F}\), \(72\,\mu\text{C}\)
ⓒ. \(9.0\,\mu\text{F}\), \(108\,\mu\text{C}\)
ⓓ. \(2.0\,\mu\text{F}\), \(24\,\mu\text{C}\)
Correct Answer: \(2.0\,\mu\text{F}\), \(24\,\mu\text{C}\)
Explanation: \( \textbf{Given:} \) \(C_1=3.0\,\mu\text{F}\), \(C_2=6.0\,\mu\text{F}\), and total voltage \(V=12\,\text{V}\).
\( \textbf{Series equivalent for two capacitors:} \)
\[
C_{\text{eq}}=\frac{C_1C_2}{C_1+C_2}
\]
\( \textbf{Substitution:} \)
\[
C_{\text{eq}}=\frac{(3.0)(6.0)}{3.0+6.0}\,\mu\text{F}
\]
\( \textbf{Simplification:} \)
\[
C_{\text{eq}}=\frac{18.0}{9.0}\,\mu\text{F}=2.0\,\mu\text{F}
\]
\( \textbf{Charge supplied to series combination:} \)
\[
Q=C_{\text{eq}}V
\]
\( \textbf{Calculation:} \)
\[
Q=(2.0\,\mu\text{F})(12\,\text{V})=24\,\mu\text{C}
\]
\( \textbf{Series condition:} \) Each capacitor carries the same charge magnitude \(24\,\mu\text{C}\).
\( \textbf{Final answer:} \) \(C_{\text{eq}}=2.0\,\mu\text{F}\) and \(Q=24\,\mu\text{C}\) on each capacitor.
211. Two capacitors \(C_1=2C\) and \(C_2=6C\) are connected in series. What is the ratio of potential differences \(V_1:V_2\)?
ⓐ. \(1:3\)
ⓑ. \(2:6\)
ⓒ. \(1:1\)
ⓓ. \(3:1\)
Correct Answer: \(3:1\)
Explanation: \( \textbf{Series condition:} \) Both capacitors carry the same charge \(Q\).
\( \textbf{Voltage relation:} \)
\[
V=\frac{Q}{C}
\]
\( \textbf{Voltage across \(C_1=2C\):} \)
\[
V_1=\frac{Q}{2C}
\]
\( \textbf{Voltage across \(C_2=6C\):} \)
\[
V_2=\frac{Q}{6C}
\]
\( \textbf{Ratio:} \)
\[
V_1:V_2=\frac{Q}{2C}:\frac{Q}{6C}
\]
\( \textbf{Simplification:} \)
\[
V_1:V_2=\frac{1}{2}:\frac{1}{6}=3:1
\]
\( \textbf{Physical meaning:} \) The smaller capacitor gets the larger share of voltage in series.
\( \textbf{Final answer:} \) \(V_1:V_2=3:1\).
212. Assertion: The equivalent capacitance of capacitors connected in series is less than the smallest individual capacitance.
Reason: In series, the reciprocal capacitances add, so \(\frac{1}{C_{\text{eq}}}\) is greater than the reciprocal of any one capacitor.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: For capacitors in series, the equivalent capacitance satisfies \(\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots\). Since all capacitances are positive, the reciprocal sum is greater than any single reciprocal term. Therefore, \(C_{\text{eq}}\) must be smaller than each individual capacitance. This is opposite to the behaviour of capacitors in parallel, where capacitances add directly. Physically, series connection increases the effective separation between conducting plates, reducing capacitance. The Reason gives the mathematical explanation of why the Assertion is true.
213. Two capacitors \(2.0\,\mu\text{F}\) and \(8.0\,\mu\text{F}\) are connected in series across a \(100\,\text{V}\) supply. What are the potential differences across the \(2.0\,\mu\text{F}\) and \(8.0\,\mu\text{F}\) capacitors?
ⓐ. \(20\,\text{V}\) and \(80\,\text{V}\)
ⓑ. \(50\,\text{V}\) and \(50\,\text{V}\)
ⓒ. \(100\,\text{V}\) and \(100\,\text{V}\)
ⓓ. \(80\,\text{V}\) and \(20\,\text{V}\)
Correct Answer: \(80\,\text{V}\) and \(20\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(C_1=2.0\,\mu\text{F}\), \(C_2=8.0\,\mu\text{F}\), and total voltage \(V=100\,\text{V}\).
\( \textbf{Series voltage division:} \) In series, charge is the same, so voltage is inversely proportional to capacitance.
\( \textbf{Ratio of voltages:} \)
\[
V_1:V_2=\frac{1}{C_1}:\frac{1}{C_2}
\]
\( \textbf{Substitution:} \)
\[
V_1:V_2=\frac{1}{2.0}:\frac{1}{8.0}=4:1
\]
\( \textbf{Total voltage:} \)
\[
V_1+V_2=100\,\text{V}
\]
\( \textbf{Share calculation:} \)
\[
V_1=\frac{4}{5}(100)=80\,\text{V}
\]
\[
V_2=\frac{1}{5}(100)=20\,\text{V}
\]
\( \textbf{Check:} \) The smaller \(2.0\,\mu\text{F}\) capacitor gets the larger voltage.
\( \textbf{Final answer:} \) The voltages are \(80\,\text{V}\) and \(20\,\text{V}\).
214. Which row correctly compares capacitors in series and in parallel?
| Row | Series capacitors | Parallel capacitors |
|---|
| P | Same charge on each | Same potential difference across each |
| Q | Same potential difference across each | Same charge on each |
| R | Equivalent capacitance is direct sum | Equivalent capacitance is reciprocal sum |
| S | Larger capacitor gets larger voltage | Smaller capacitor gets larger charge |
ⓐ. Row S
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row P
Correct Answer: Row P
Explanation: In a series combination, the same charge magnitude appears on each capacitor. In a parallel combination, each capacitor is connected across the same two terminals, so the same potential difference appears across each capacitor. Row Q reverses these rules. Row R is also reversed because series capacitors combine by reciprocal addition, while parallel capacitors add directly. Row S is wrong because in series the smaller capacitor gets larger voltage, and in parallel the larger capacitor gets larger charge at the same voltage. The safest way to remember the contrast is: series means same \(Q\), parallel means same \(V\).
215. A graph is plotted for capacitors connected in series, all carrying the same charge \(Q\). The vertical axis is potential difference \(V\) across a capacitor, and the horizontal axis is \(\frac{1}{C}\). What is the slope of the graph?
ⓐ. \(\frac{1}{Q}\)
ⓑ. \(C_{\text{eq}}\)
ⓒ. \(Q\)
ⓓ. \(Q^2\)
Correct Answer: \(Q\)
Explanation: \( \textbf{Series condition:} \) Each capacitor in series carries the same charge magnitude \(Q\).
\( \textbf{Voltage relation:} \)
\[
V=\frac{Q}{C}
\]
\( \textbf{Graph variable:} \) The horizontal axis is \(\frac{1}{C}\).
\( \textbf{Straight-line form:} \)
\[
V=Q\left(\frac{1}{C}\right)
\]
\( \textbf{Slope comparison:} \) Comparing with \(y=mx\), \(V\) is \(y\), \(\frac{1}{C}\) is \(x\), and \(Q\) is the slope.
\( \textbf{Physical meaning:} \) A capacitor with smaller \(C\) has larger \(\frac{1}{C}\), so it gets larger voltage for the same charge.
\( \textbf{Final answer:} \) The slope of the graph is \(Q\).
216. Two capacitors \(C_1=3.0\,\mu\text{F}\) and \(C_2=6.0\,\mu\text{F}\) are connected in series across \(18\,\text{V}\). What are the voltages across \(C_1\) and \(C_2\)?
ⓐ. \(12\,\text{V}\) and \(6\,\text{V}\)
ⓑ. \(9\,\text{V}\) and \(9\,\text{V}\)
ⓒ. \(6\,\text{V}\) and \(12\,\text{V}\)
ⓓ. \(18\,\text{V}\) and \(18\,\text{V}\)
Correct Answer: \(12\,\text{V}\) and \(6\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(C_1=3.0\,\mu\text{F}\), \(C_2=6.0\,\mu\text{F}\), and total voltage \(18\,\text{V}\).
\( \textbf{Series condition:} \) Charge \(Q\) is the same on both capacitors.
\( \textbf{Voltage relation:} \)
\[
V=\frac{Q}{C}
\]
\( \textbf{Voltage ratio:} \)
\[
V_1:V_2=\frac{1}{3.0}:\frac{1}{6.0}=2:1
\]
\( \textbf{Total voltage condition:} \)
\[
V_1+V_2=18\,\text{V}
\]
\( \textbf{First voltage:} \)
\[
V_1=\frac{2}{3}(18)=12\,\text{V}
\]
\( \textbf{Second voltage:} \)
\[
V_2=\frac{1}{3}(18)=6\,\text{V}
\]
\( \textbf{Final answer:} \) \(V_1=12\,\text{V}\) and \(V_2=6\,\text{V}\).
217. In a series combination, two capacitors have capacitances \(C\) and \(3C\). Which capacitor stores more energy?
ⓐ. Both store equal energy
ⓑ. Energy is zero in both because the charges are equal
ⓒ. The capacitor \(C\) stores more energy
ⓓ. The capacitor \(3C\) stores more energy
Correct Answer: The capacitor \(C\) stores more energy
Explanation: In series, both capacitors carry the same charge \(Q\). For a capacitor with fixed charge, the stored energy is \(U=\frac{Q^2}{2C}\). This expression shows that, for the same \(Q\), energy is inversely proportional to capacitance. The smaller capacitor \(C\) stores more energy than the larger capacitor \(3C\). The larger capacitor has smaller voltage and smaller energy under the same-charge condition. A common misconception is to assume larger capacitance always stores more energy; that is true at fixed voltage, not at fixed charge.
218. Two capacitors \(4.0\,\mu\text{F}\) and \(12.0\,\mu\text{F}\) are connected in series. The maximum safe voltage across the \(4.0\,\mu\text{F}\) capacitor is \(60\,\text{V}\), and across the \(12.0\,\mu\text{F}\) capacitor is \(30\,\text{V}\). What is the maximum safe total voltage for the series combination?
ⓐ. \(120\,\text{V}\)
ⓑ. \(80\,\text{V}\)
ⓒ. \(90\,\text{V}\)
ⓓ. \(40\,\text{V}\)
Correct Answer: \(80\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(C_1=4.0\,\mu\text{F}\), \(C_2=12.0\,\mu\text{F}\), safe limit for \(C_1\) is \(60\,\text{V}\), and safe limit for \(C_2\) is \(30\,\text{V}\).
\( \textbf{Series condition:} \) Both capacitors carry the same charge \(Q\).
\( \textbf{Voltage ratio:} \)
\[
V_1:V_2=\frac{1}{C_1}:\frac{1}{C_2}
\]
\( \textbf{Substitution:} \)
\[
V_1:V_2=\frac{1}{4.0}:\frac{1}{12.0}=3:1
\]
\( \textbf{Limit check:} \) If \(V_1=60\,\text{V}\), then \(V_2=20\,\text{V}\), which is below its \(30\,\text{V}\) limit.
\( \textbf{If \(V_2=30\,\text{V}\):} \) Then \(V_1=90\,\text{V}\), which exceeds the \(60\,\text{V}\) limit.
\( \textbf{Allowed maximum:} \)
\[
V_{\text{total}}=60\,\text{V}+20\,\text{V}=80\,\text{V}
\]
\( \textbf{Final answer:} \) The maximum safe total voltage is \(80\,\text{V}\).
219. Consider the following statements about capacitors in series.
Statement I: The same charge magnitude appears on each capacitor.
Statement II: The equivalent capacitance is smaller than the smallest individual capacitance.
Statement III: The largest capacitance gets the largest potential difference.
Which statements are correct?
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is correct because capacitors in series carry the same charge magnitude. Statement II is correct because reciprocal capacitances add in series, making \(C_{\text{eq}}\) smaller than the smallest capacitor. Statement III is incorrect because voltage across a series capacitor is \(V=\frac{Q}{C}\). Since \(Q\) is the same, the smaller capacitance gets the larger potential difference. The voltage distribution is therefore inverse to capacitance, not directly proportional to it. A common series-combination mistake is to think the largest capacitor must get the largest share of voltage.
220. Two capacitors \(C\) and \(2C\) are connected first in parallel and then in series. What is the ratio of equivalent capacitance in parallel to equivalent capacitance in series?
ⓐ. \(\frac{9}{2}:1\)
ⓑ. \(3:1\)
ⓒ. \(1:\frac{9}{2}\)
ⓓ. \(2:3\)
Correct Answer: \(\frac{9}{2}:1\)
Explanation: \( \textbf{Parallel equivalent:} \)
\[
C_{\text{p}}=C+2C=3C
\]
\( \textbf{Series equivalent:} \)
\[
C_{\text{s}}=\frac{C(2C)}{C+2C}
\]
\( \textbf{Simplification of series value:} \)
\[
C_{\text{s}}=\frac{2C^2}{3C}=\frac{2C}{3}
\]
\( \textbf{Required ratio:} \)
\[
C_{\text{p}}:C_{\text{s}}=3C:\frac{2C}{3}
\]
\( \textbf{Cancel \(C\):} \)
\[
C_{\text{p}}:C_{\text{s}}=3:\frac{2}{3}
\]
\( \textbf{Final simplification:} \)
\[
C_{\text{p}}:C_{\text{s}}=\frac{9}{2}:1
\]
\( \textbf{Final answer:} \) The ratio is \(\frac{9}{2}:1\).
