Combination of two identical capacitors, a resistor R and a DC voltage source of voltage 6 V is used

Q: Why do we use the discharging equation instead of the charging one?

The problem states that the voltage of the *fully charged* combination reduces to half its original voltage. This implies the capacitors are losing their stored energy through the resistor, which is the definition of discharging.

Q: Does the initial voltage of 6 V affect the time taken to reach half voltage?

Not at all! The time it takes for an RC circuit to discharge to any fraction of its initial voltage depends only on the time constant $\tau = RC$, not on the initial voltage itself. The $V_0$ term cancels out in the equation.

Q: Why is the equivalent capacitance in series $C/2$ and not $2C$?

In series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances: $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$. Inverting this gives $C_{eq} = \frac{C}{2}$.

Q: What is the significance of the term $RC \ln 2$?

This is the 'half-life' of the RC circuit. Just like in radioactive decay, it's the exact time required for the voltage, charge, or current to drop to 50% of its initial value.

Q: Why does the series combination discharge faster?

The series combination has a smaller equivalent capacitance ($C/2$ compared to $2C$). A smaller capacitance means it stores less charge for the same voltage. With less charge to push through the same resistor, it empties out much faster!

Question

Combination of two identical capacitors, a resistor $R$ and a DC voltage source of voltage 6 V is used in an experiment on $C-R$ circuit. It is found that for a parallel combination of the capacitor, the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 s. For series combination, the time needed for reducing the voltage of the fully charged series combination by half is

Options:

(A) 20 s

(B) 10 s

(C) 5 s

(D) 2.5 s

Sigma Percentile · Accepted Answer

Imagine you are an engineer tasked with designing a timing circuit. You have two identical capacitors, a resistor, and a 6 V battery. You decide to run a little experiment to see how the arrangement of the capacitors affects the time it takes for them to discharge. This problem is a classic exploration of RC circuits and the concept of the time constant.

### Analyzing the Setup

First, let's understand what happens when a fully charged capacitor discharges through a resistor. The voltage doesn't drop instantly; instead, it decays exponentially over time. Think of a capacitor as a water tank and the resistor as a narrow pipe. When the valve is opened, the water rushes out quickly at first because the pressure is high. But as the tank empties, the pressure drops, and the flow slows down.

The mathematical equation governing this exponential decay is:

$$V(t) = V_0 e^{-\frac{t}{R C_{eq}}}$$

Here, $V_0$ is the initial voltage (6 V in our case), $R$ is the resistance, and $C_{eq}$ is the equivalent capacitance of the circuit. The product $R C_{eq}$ is known as the **time constant** ($	au$), which dictates how sluggishly or rapidly the circuit discharges.

### The Master Equation

The problem asks us to find the time it takes for the voltage to reduce to exactly half of its original value. Let's set $V(t) = \frac{V_0}{2}$ and solve for $t$:

$$\frac{V_0}{2} = V_0 e^{-\frac{t}{R C_{eq}}}$$

The $V_0$ beautifully cancels out on both sides, leaving us with:

$$\frac{1}{2} = e^{-\frac{t}{R C_{eq}}}$$

Taking the natural logarithm on both sides, we arrive at the master equation for the half-life of an RC circuit:

$$t = R C_{eq} \ln 2$$

This tells us that the time to reach half voltage is directly proportional to the equivalent capacitance $C_{eq}$.

### The Parallel Marathon

In the first part of the experiment, the two identical capacitors (each of capacitance $C$) are connected in parallel. When capacitors are in parallel, it's like having two water tanks side by side. The total capacity is huge! The equivalent capacitance is simply the sum:

$$C_{eq, p} = C + C = 2C$$

Plugging this into our master equation, the time $t_1$ for the parallel combination to discharge to half voltage is:

$$t_1 = R (2C) \ln 2$$

The problem states that this takes exactly 10 seconds. So, $R (2C) \ln 2 = 10	ext{ s}$.

### The Series Sprint

Now, let's look at the series combination. When capacitors are connected in series, the effective capacity is actually reduced. It's like stacking the tanks in a weird way where the overall ability to store charge drops. The equivalent capacitance is:

$$C_{eq, s} = \frac{C 	imes C}{C + C} = \frac{C}{2}$$

The time $t_2$ for the series combination to reach half voltage is:

$$t_2 = R \left(\frac{C}{2}ight) \ln 2$$

### Final Calculation

To find $t_2$, we can simply take the ratio of the two times:

$$\frac{t_2}{t_1} = \frac{R (C/2) \ln 2}{R (2C) \ln 2}$$

Notice how the $R$, $C$, and $\ln 2$ terms cancel out perfectly:

$$\frac{t_2}{t_1} = \frac{1/2}{2} = \frac{1}{4}$$

This means the series combination discharges four times faster than the parallel combination! Since $t_1 = 10	ext{ s}$, we can easily find $t_2$:

$$t_2 = \frac{10}{4} = 2.5	ext{ s}$$

And there we have it! By simply rearranging the capacitors from parallel to series, we slashed the discharge time from 10 seconds down to a mere 2.5 seconds. The physics of RC circuits is truly elegant.

Select Answer:

Visualized Solution

Visualizing the Two Circuits

The Discharging Equation

Deriving the Half-Life Formula

Time for Parallel Combination

Time for Series Combination

Calculating the Final Ratio

The Way Forward

The Sigma Insight: RC Circuit

Analyzing the Setup

The Master Equation

The Parallel Marathon

The Series Sprint

Final Calculation

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