You visited us 1 times! Enjoying our articles? Unlock Full Access!

Potentiometer

The switch in...

Question

The switch in circuit shifts from $1$ to $2$ when $V_{C} > 2 V / 3$ and goes back to $1$ from $2$ when $V_{C} < 2 V / 3$ . The voltage across capacitor $V_{C}$ as read by voltmeter is plotted. What is the period $T$ of the waveform in terms of $R$ and $C$ ?

R C ln 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 R C ln 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{R C}{2} ln 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R C}{3} ln 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 R C ln 2$
During time interval $t_{2}$ , the capacitor is discharging with the help of resistor $R$ because the potential difference across it is decreasing $(q \propto V)$

$\Rightarrow q = q_{0} e^{- t_{2} / R C}$

$\Rightarrow C V = C V_{0} e^{- t_{2} / R C}$

$\Rightarrow V = V_{0} e^{- t_{2} / R C}$

From the graph,

$V_{0} = \frac{2 V}{3}$ and $V = \frac{V}{3}$

$\Rightarrow \frac{2 V}{3} e^{- t_{2} / R C} = \frac{V}{3}$

$\Rightarrow 2 = e^{t_{2} / R C}$

Taking $ln$ both sides,

$\Rightarrow ln 2 = \frac{t_{2}}{R C}$

$\Rightarrow t_{2} = (ln 2) R C$

During time interval $t_{1}$ , the capacitor is charging with battery ( $∵ V ↑$ ). So, the charging equation,

$q = q_{0} (1 - e^{- t_{1} / R C})$

$\Rightarrow C V = C V_{0} (1 - e^{- t_{1} / R C})$

$\Rightarrow V = V_{0} (1 - e^{- t_{1} / R C})$

Again from the graph,

$\frac{V}{3} = \frac{2 V}{3} (1 - e^{- t_{1} / R C})$

$\Rightarrow \frac{1}{2} = 1 - e^{- t_{1} / R C}$

$\Rightarrow e^{- t_{1} / R C} = \frac{1}{2}$

$\Rightarrow e^{t_{1} / R C} = 2$

Taking $ln$ both sides, we get

$t_{1} = R C ln 2$

Thus the period of wave form is;

$T = t_{1} + t_{2}$

$\Rightarrow T = 2 R C ln 2$

Hence, option $(B)$ is the correct answer.

Suggest Corrections

Similar questions

In the given R – C circuit switch S is closed at t = 0 and the variation of potential drop across the resistor $(V_{R})$ and across capacitor $(V_{C})$ is represented by given plot, then :

Q. For the circuit shown below,

the voltage across the capacitor

(V_{c})

Q. When the rms voltages

V_{L}, V_{C}

and

V_{R}

are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio

V_{L} : V_{C} : V_{R} = 1 : 2 : 3

. If the rms voltage of the AC source is 100 V, then

V_{R}

is close to :

Q. In an LCR circuit, voltages across the components are

V_{L}

V_{C}

V_{R}

respectively when connected to a AC source

V = V_{0} s i n (ω t)

. The relation between

V

V_{R}

V_{C}

and

V_{L}

is.

Q. The switch has been in position 1 for a long time and abruptly changes to position 2 at t = 0.

If time t is in seconds, the capacitor voltage

V_{C}

(in volts) for t > 0 is given by

Join BYJU'S Learning Program

Explore more

Potentiometer

Standard XII Physics

Join BYJU'S Learning Program

The switch in circuit shifts from 1 to 2 when VC>2V/3 and goes back to 1 from 2 when VC<2V/3. The voltage across capacitor VC as read by voltmeter is plotted. What is the period T of the waveform in terms of R and C ?

The switch in circuit shifts from $1$ to $2$ when $V_{C} > 2 V / 3$ and goes back to $1$ from $2$ when $V_{C} < 2 V / 3$ . The voltage across capacitor $V_{C}$ as read by voltmeter is plotted. What is the period $T$ of the waveform in terms of $R$ and $C$ ?