Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given ca

Q: Why is the slope equal to capacitance?

Because $Q=CV$, so $Q/V=C$, which is exactly the definition of the slope of a $Q$ vs $V$ graph.

Q: How do we know Line A is the parallel combination?

Parallel combination always yields a higher equivalent capacitance than series. Since Line A has a steeper slope, it represents a higher capacitance.

Q: Can we solve this without forming a quadratic equation?

Yes! Once you have $C_1+C_2=50$ and $C_1C_2=400$, you can simply guess the two numbers that satisfy both conditions. $40$ and $10$ are the obvious choices.

Q: What if the axes were swapped ($V$ vs $q$)?

Then the slope would be $1/C$. In that case, the line with the lower slope would represent the parallel combination.

Q: Are the units important here?

Absolutely. The y-axis is in $\mu\text{C}$ and the x-axis is in Volts. Dividing them gives the capacitance in $\mu\text{F}$.

Question

Figure shows charge ($q$) versus voltage ($V$) graph for series and parallel combination of two given capacitors. The capacitances are

Options:

(A) $60\mu	ext{F}$ and $40\mu	ext{F}$

(B) $50\mu	ext{F}$ and $30\mu	ext{F}$

(C) $20\mu	ext{F}$ and $30\mu	ext{F}$

(D) $40\mu	ext{F}$ and $10\mu	ext{F}$

Sigma Percentile · Accepted Answer

### Decoding the Graph

Imagine you are looking at a map, but instead of distance and time, it shows charge ($q$) and voltage ($V$). What does the slope of this map tell us?

We know the fundamental equation for a capacitor is:

$$Q = CV$$

If we rearrange this, we get $C = \frac{Q}{V}$. This is exactly the definition of the slope of a $q-V$ graph! So, a steeper line means a higher capacitance.

### Identifying the Combinations

The problem gives us two lines, $A$ and $B$, representing the series and parallel combinations of two capacitors, $C_1$ and $C_2$.

Think about it: which combination gives a higher equivalent capacitance?

**Parallel combination** always results in a higher equivalent capacitance ($C_P = C_1 + C_2$) compared to the **series combination** ($C_S = \frac{C_1 C_2}{C_1 + C_2}$).

Therefore, the steeper line, **Line A**, must represent the parallel combination, and the less steep line, **Line B**, represents the series combination.

### Extracting Data from the Graph

Let's extract the exact values from the graph.

For **Line A** (Parallel Combination):
At a voltage of $10	ext{ V}$, the charge is $500\mu	ext{C}$.

$$C_P = \frac{500\mu	ext{C}}{10	ext{ V}} = 50\mu	ext{F}$$

This gives us our first equation:

$$C_1 + C_2 = 50$$

For **Line B** (Series Combination):
At a voltage of $10	ext{ V}$, the charge is $80\mu	ext{C}$.

$$C_S = \frac{80\mu	ext{C}}{10	ext{ V}} = 8\mu	ext{F}$$

This gives us our second equation:

$$\frac{C_1 C_2}{C_1 + C_2} = 8$$

### The Master Equation

Now, we have a system of two equations. Let's substitute the value of $C_1 + C_2$ from the first equation into the second one.

$$\frac{C_1 C_2}{50} = 8$$

$$C_1 C_2 = 400$$

We need to find two numbers that add up to $50$ and multiply to $400$. We can set up a quadratic equation by substituting $C_1 = 50 - C_2$:

$$C_2(50 - C_2) = 400$$

$$50C_2 - C_2^2 = 400$$

Rearranging this into a standard quadratic form:

$$C_2^2 - 50C_2 + 400 = 0$$

### Final Calculation

I know this quadratic equation looks terrifying, but let's take a breath. We just need to factorize it. The numbers $40$ and $10$ work perfectly!

$$(C_2 - 40)(C_2 - 10) = 0$$

This gives us two possible values for $C_2$:

$$C_2 = 40\mu	ext{F} \quad 	ext{or} \quad C_2 = 10\mu	ext{F}$$

If $C_2 = 40\mu	ext{F}$, then $C_1 = 10\mu	ext{F}$. 
If $C_2 = 10\mu	ext{F}$, then $C_1 = 40\mu	ext{F}$.

In either case, the two capacitors have capacitances of **$40\mu	ext{F}$ and $10\mu	ext{F}$**.

Animated Solution for Physics - Electrostatics: Figure shows charge ( $q$ ) versus voltage ( $V$ ) graph for series and parallel combination of two given capacitors. The capacitances are

Select Answer:

Visualized Solution

$Q - V Graph$

$Identifying Combinations$

$Parallel Capacitance C_{P}$

$Series Capacitance C_{S}$

$Solving for C_{1} and C_{2}$

$Quadratic Equation$

$Final Values$

The Sigma Insight: Combination of Capacitors

Decoding the Graph

Identifying the Combinations

Extracting Data from the Graph

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q−V Graph

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Combinations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Capacitance CP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Series Capacitance CS​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for C1​ and C2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Quadratic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Values