A 4 microfarad capacitor and a resistance of 2.5 M ohm are in series with 12 V battery. Find the tim

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

The problem states the capacitor and resistor are in series with a battery, which implies the capacitor is being charged from an initially uncharged state.

Q: What happens to the voltage across the resistor as time passes?

As the capacitor charges, it opposes the current flow. The current decreases exponentially, and since $V_R = IR$, the voltage across the resistor also decreases exponentially.

Q: Why does the battery voltage $V$ cancel out?

Both $V_C$ and $V_R$ are proportional to the applied battery voltage $V$. When we take their ratio or equate them with a multiplier, the source voltage $V$ factors out, meaning the time depends only on the RC time constant.

Q: Can the voltage across the capacitor ever exceed the battery voltage?

No, in a simple DC RC circuit, the maximum voltage the capacitor can reach is exactly equal to the battery voltage, which happens theoretically at $t = \infty$.

Q: What is the significance of the time constant $\tau$?

The time constant $\tau = RC$ represents the time it takes for the capacitor to charge to approximately 63.2% of its maximum voltage. It dictates the speed of the charging process.

Question

A $4\mu	ext{F}$ capacitor and a resistance of $2.5	ext{ M}\Omega$ are in series with $12	ext{ V}$ battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given, $\ln (2) = 0.693$]

Options:

(A) 13.86 s

(B) 6.93 s

(C) 7 s

(D) 14 s

Sigma Percentile · Accepted Answer

The problem asks us to find the exact moment when the voltage across a charging capacitor becomes three times the voltage across the resistor in a simple RC circuit.

### Analyzing the Setup

Imagine you have a $12	ext{ V}$ battery connected in series with a $4\mu	ext{F}$ capacitor and a $2.5	ext{ M}\Omega$ resistor. When the circuit is closed, the battery starts pumping charge onto the capacitor plates.

Initially, the capacitor acts like a short circuit, and all the voltage drops across the resistor. But as time passes, the capacitor fills up with charge, and its voltage $V_C$ increases, while the current in the circuit drops, causing the resistor's voltage $V_R$ to decrease.

### The Master Equation

The voltage across the capacitor during charging is given by the exponential growth equation:
$$V_C = V(1 - e^{-t/	au})$$

Simultaneously, the voltage across the resistor decays exponentially:
$$V_R = V e^{-t/	au}$$

The problem gives us a fascinating constraint: we need to find the time $t$ when $V_C = 3 V_R$. Let's substitute our formulas into this condition:
$$V(1 - e^{-t/	au}) = 3 V e^{-t/	au}$$

Notice how the battery voltage $V$ appears on both sides? It completely cancels out! This means the time it takes to reach this voltage ratio is entirely independent of the battery you use.

$$1 - e^{-t/	au} = 3 e^{-t/	au}$$
$$1 = 4 e^{-t/	au}$$
$$e^{t/	au} = 4$$

Taking the natural logarithm on both sides, we get:
$$\frac{t}{	au} = \ln(4) = 2 \ln(2)$$

### Final Calculation

Before we find $t$, we need the time constant $	au$, which dictates how fast the circuit charges. 
$$	au = RC$$
$$	au = (2.5 	imes 10^6 \Omega) 	imes (4 	imes 10^{-6} 	ext{ F}) = 10 	ext{ s}$$

Now, we just plug this back into our time equation. We are given $\ln(2) = 0.693$.
$$t = 10 	imes 2 	imes 0.693$$
**$$t = 13.86 	ext{ s}$$**

And there we have it! After exactly $13.86$ seconds, the capacitor will hold three times the voltage of the resistor.

Animated Solution for Physics - Current Electricity: A $4 μ F$ capacitor and a resistance of $2.5 M Ω$ are in series with $12 V$ battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given, $ln (2) = 0.693$ ]

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Visualized Solution

$R C Circuit Setup$

$Charging Formulas$

$Given Condition$

$Substitution$

$Simplification$

$Logarithm$

$Time Constant$

$Final Calculation$

The Sigma Insight: RC Circuit

Analyzing the Setup

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; } RC Circuit Setup

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

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

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

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

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

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

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