RC Circuits MCQ Quiz in मल्याळम - Objective Question with Answer for RC Circuits - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is option '1'.

Concept:

• In an R-C circuit the current leads voltage.
• The voltage applied to an R-C circuit is equal to vector sum of voltage across resistor and capacitor i.e., 

 .$V=V_R-i{V}_C$

In magnitude

$V^2=V_R^2+V_C^2$

Calculation:

V_{R =} 12 V 

V_C = 16 V

Let V be the total voltage applied across R-C circuit. So, V equals to

V² = 12² + 16²

V² = 400

V = 20 V

Concept:

In an AC circuit, the voltages across the resistor (heater) and the capacitor are not in phase. Therefore, we cannot simply add them algebraically. Instead, we use the following relationship, derived from the Pythagorean theorem:

Explanation:

Given:
Supply voltage, $V_{\text{total}}=220\text{V}$
Voltage across the heater (resistor), $V_R=90\text{V}$
Frequency of AC supply, $f=50\text{Hz}$

The potential difference across the capacitor, $V_C$

The total potential for RC series circuit is given by:

$V_{\text{total}}^2=V_R^2+V_C^2$

where:
$V_{\text{total}}$ is the total supply voltage,
$V_R$ is the voltage across the resistor,
$V_C$ is the voltage across the capacitor.

Substituting the given values:

$$\begin{gathered} (220{)}^2=(90{)}^2+V_C^2 \\ 48400=8100+V_C^2 \\ {V}_C^2=48400-8100=40300 \\ {V}_C=\sqrt{40300}\approx 200.75\text{V} \end{gathered}$$

The correct option is (1).
 

CONCEPT:

The capacitance of the capacitor is written as;

Q = VC

Here we have Q as the charge, V as the potential, and C as the capacitance of the capacitor.

CALCULATION:

Given: Capacitance of capacitor, C = 4 μF 

Internal resistance, R = 0.5 Ω 

At the initial point, the capacitor gets charged, and when it gets charged no current will flow through the circuit.

As we know;

V = IR

$I=\frac{V}{R}$

Now, on putting all the given values we have;

⇒$I=\frac{2.5}{2+0.5}$

⇒I = 1 A

Now, the voltage in the capacitor is written as;

V_c = I × R = 2 V

and the charge on the capacitors is written as;

qc = V' × C

⇒ qc = 2 × 4 μC = 8 μC

Hence, option 4) is the correct answer.

CONCEPT:

The growth of the charge on the capacitor is written as;

$Q=Q_o(1-e^{-\frac{t}{RC}})$

EXPLANATION:

As we have;

$Q=Q_o(1-e^{-\frac{t}{RC}})$

from the equation of charge on the capacitor, it can be seen that the product of C and R decreases, e^-t/RC decreases, and the value charge $Q=Q_o(1-e^{-\frac{t}{RC}})$increases.

If CR is small in the given equation, then the time constant is also small and the growth of the charge will more rapid.

Concept:

• The combination of a pure resistance R in ohms and pure capacitance C in Farads is called an RC circuit.
• The capacitor stores energy and the resistor connected in series with the capacitor controls the charging and discharging of the capacitor.
• The RC circuit is used in camera flashes, pacemakers, timing circuits etc.
• $I=\frac{V}{\sqrt{R^2+X_C^2}}$
• $X_c=\frac{1}{\omega C}$
• Where, R = resistance, X_c = reactance, V = voltage. I = current, ω = angular frequency, 

​Calculation:

Given, Initial reactance at angular frequency ω is $X_c=\frac{1}{\omega C}$

Then $I=\frac{V}{\sqrt{R^2+X_C^2}}$ --- (1)

Final reactance at angular frequency $\frac{\omega }{3}$ is $X_{cf}=\frac{3}{\omega C}=3X_C$

And $I^{\prime }=\frac{I}{2}$ , voltage is same

Then, $\frac{I}{2}=\frac{V}{\sqrt{R^2+9X_C^2}}$ ---- (2)

Dividing equation 1 by 2 then

$2=\frac{\sqrt{R^2+9X_C^2}}{\sqrt{R^2+X_C^2}}$

⇒ 4(R² + X_C²) = R² + 9X_C²

⇒ $\frac{X_C}{R}=\sqrt{\frac{3}{5}}$