**From the given Figure, answer the questions either True or False based on the circuit’s behavior:**

**– After the RELAY is switched to either the N.O. (“normally open”) or N.C. (“normally closed”) state, the transient response of the circuit is for the short time.**

**– In this experiment, the transient current flow has an exponential decay to zero.**

**– The charge Q of the capacitor decay exponentially when the relay move to the N. O. state.**

**– The capacitor charge Q decreases during current I is positive.**

**– The negative voltage measured in VOLTAGE IN 2 is due to positive current I.**

**– VOLTAGE IN 1 is measured to be positive when the charge Q on the capacitor is positive. **

**– The given quantity t1/2=? ln 2 is the half-life of an exponential decay, where ?= R.C. is the time constant in an R.C. circuit. The current in a discharging R.C. circuit drops by half whenever t increases by $t_{12}$. For a circuit with $R=2k\Omega$ and $C=3uF$, if at t=5 ms the current is 6 mA, find the time (in ms) that the current of 3 mA would be.**

This question aims to find the **current, charge, and voltage** in the **RC circuit**. There are multiple statements given and the task is to find the correct one.

Moreover, this question is based on the concepts of physics. In the **RC circuit**, the **capacitor** is charged when it is connected to the source. However, when the source is disconnected, the **capacitor** discharges through the **resistor**.

## Expert Answer

1) As the **capacitor** is initially uncharged, it resists the change in **voltage** instantaneously. Hence,

Voltage, when the switch is closed the initial current,

\[ i =\dfrac{V_s}{R} \]

So, the statement is true.

2) At any instant the current is:

\[ i =\dfrac{(V_s – V_c)}{R} \]

Furthermore, the increase in **voltage** causes the $i=0$, therefore:

\[ V_c = V_s \]

So, the statement is true.

3) When $V_s$ is connected, the voltage across a capacitor **increases exponentially** till it reaches a steady state. Therefore, the charge is:

\[q = CV_s\]

So, the statement is false.

4) The direction of current shown in the figure prove that the charge in the capacitor is increasing.

So, the statement is false.

5) The **voltage** across the **capacitor** and the resistor is positive, therefore, Voltage IN 2 will be positive.

So, the statement is false.

6) According to **Kirchoff voltage law**, Voltage OUT 1 and Voltage IN 1 are equal.

So, the statement is false.

7) The **capacitor’s current** equation is:

\[I(t) = \dfrac{V_s}{R}[1 -\exp(-t/RC)]\]

Since,

$I=6mA$

$t=5ms$

Therefore,

\[\dfrac{V_s}{R}=10.6mA\]

\[3 mA = 10.6 mA [1 – \exp(-t/(2k\Omega \times 3uF) )]\]

\[\Rightarrow t=2ms\]

## Numerical Results

**current**is 3mA is:

\[t=2ms\]

## Example

When the current through a 10k\Omega resistor is 5mA, find the voltage against it.

**Solution:**

The voltage can be found as:

\[V = IR = 5mA \times 10k\Omega\]

\[V = 50V\]

*Images/Mathematical are created with Geogebra.*