At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the cap

Q: Why are the two resistors in parallel?

Because both resistors are connected across the same two nodes. The current from terminal A splits into two paths, goes through the resistors, and then recombines at the vertical wire.

Q: Why do we add the capacitances for parallel capacitors?

Unlike resistors, capacitors in parallel have their equivalent capacitance calculated by simply adding them up. This is because the effective plate area increases, allowing them to store more charge for the same voltage.

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

The time constant $\tau = RC$ represents the time it takes for the capacitor to charge to about 63.2% of its maximum voltage. It dictates how fast or slow the circuit responds.

Q: Why do we use the equation $V_c(t) = V_0(1 - e^{-t/\tau})$?

This is the fundamental solution to the differential equation governing an RC charging circuit. It shows that the voltage starts at zero and exponentially approaches the battery voltage $V_0$.

Q: How did we get $\ln(3/5) = \ln 3 - \ln 5$?

This is a standard property of logarithms: the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator, i.e., $\ln(a/b) = \ln a - \ln b$.

Question

At time $t = 0$, a battery of 10 V is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become 4 V? [Take : $\ln 5 = 1.6$, $\ln 3 = 1.1$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an electron standing at terminal A, looking at the path ahead. You see the circuit splitting into two parallel branches, each containing a resistor, and then recombining before splitting again into two parallel capacitors.

This might look intimidating at first glance, but it is actually a classic RC circuit in disguise. Our first goal is to simplify this complex web into a single equivalent resistor and a single equivalent capacitor.

### Simplifying the Circuit

Let's tackle the resistors first. We have two $2	ext{M}\Omega$ resistors connected in parallel.

When resistors are in parallel, their equivalent resistance is given by the product over sum rule. Since they are identical, the equivalent resistance is simply half of one resistor.

$$R_{	ext{eq}} = \frac{2 	imes 2}{2 + 2} = 1	ext{M}\Omega$$

Next, we look at the capacitors. We have two $2\mu	ext{F}$ capacitors, also in parallel.

Unlike resistors, capacitors in parallel add up directly. This is because connecting them in parallel effectively increases the total plate area, allowing them to store more charge.

$$C_{	ext{eq}} = 2 + 2 = 4\mu	ext{F}$$

### The Master Equation

Now, our circuit is beautifully simplified to a single $1	ext{M}\Omega$ resistor in series with a $4\mu	ext{F}$ capacitor.

The behavior of this circuit is governed by its time constant, $	au$, which dictates how quickly the capacitor charges.

$$	au = R_{	ext{eq}} C_{	ext{eq}} = (1 	imes 10^6) 	imes (4 	imes 10^{-6}) = 4	ext{s}$$

When a battery of voltage $V_0$ is connected, the voltage across the capacitor $V_c(t)$ as a function of time is given by the standard charging equation:

$$V_c(t) = V_0 (1 - e^{-t/	au})$$

### Final Calculation

We are given that the battery voltage $V_0$ is $10	ext{V}$, and we want to find the time $t$ when the voltage across the capacitor reaches $4	ext{V}$.

Let's substitute our known values into the master equation:

$$4 = 10 (1 - e^{-t/4})$$

Dividing both sides by 10, we get:

$$0.4 = 1 - e^{-t/4}$$

Rearranging the terms to isolate the exponential part:

$$e^{-t/4} = 1 - 0.4 = 0.6 = \frac{3}{5}$$

To solve for $t$, we take the natural logarithm on both sides. Remember the property of logarithms: $\ln(a/b) = \ln a - \ln b$.

$$-\frac{t}{4} = \ln\left(\frac{3}{5}ight) = \ln 3 - \ln 5$$

The problem kindly provides the values $\ln 3 = 1.1$ and $\ln 5 = 1.6$. Let's plug them in:

$$-\frac{t}{4} = 1.1 - 1.6 = -0.5$$

Multiplying both sides by $-4$, we arrive at our final answer:

$$t = 4 	imes 0.5 = 2	ext{s}$$

And there we have it! The voltage across the capacitors will reach $4	ext{V}$ at exactly $2	ext{s}$.

Animated Solution for Physics - Current Electricity: At time $t = 0$ , a battery of 10 V is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become 4 V? [Take : $ln 5 = 1.6$ , $ln 3 = 1.1$ ]

Enter Numerical Value:

Visualized Solution

$Circuit Analysis$

$R_{eq} Calculation$

$C_{eq} Calculation$

$Time Constant (τ)$

$Voltage Equation$

$Solving for t$

$Applying Logarithm$

$Final Calculation$

$Conclusion$

The Sigma Insight: RC Circuit

Analyzing the Setup

Simplifying the Circuit

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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