Time Constant MCQ Quiz - Objective Question with Answer for Time Constant - Download Free PDF

Concept:

The time constant (τ) of an RC circuit is given by the formula:

τ = R_eqC

where R_eq is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

R_eq = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = R_eqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Concept:

First-order system:

The transfer function of the standard first-order system is given by

\(TF = \frac{{k}}{{\left( {1 + τ s} \right)}}\)

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}\)

\( = \frac{{3K}}{{2s + 3K + 1}}\)

\( = \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}\)

From the above transfer function, the time constant is

\(T = \frac{2}{{3K + 1}}\)

The time constant of the system is less than 0.2

\( \Rightarrow \frac{2}{{3K + 1}} < 0.2\;\)

⇒ K > 3   

Concept:

First-order system:

The transfer function of the standard first-order system is given by

\(TF = \frac{{k}}{{\left( {1 + τ s} \right)}}\)

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}\)

\( = \frac{{3K}}{{2s + 3K + 1}}\)

\( = \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}\)

From the above transfer function, the time constant is

\(T = \frac{2}{{3K + 1}}\)

The time constant of the system is less than 0.2

\( \Rightarrow \frac{2}{{3K + 1}} < 0.2\;\)

⇒ K > 3   

Concept:

The time constant (τ) of an RC circuit is given by the formula:

τ = R_eqC

where R_eq is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

R_eq = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = R_eqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Concept:

First-order system:

The transfer function of the standard first-order system is given by

\(TF = \frac{{k}}{{\left( {1 + τ s} \right)}}\)

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}\)

\( = \frac{{3K}}{{2s + 3K + 1}}\)

\( = \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}\)

From the above transfer function, the time constant is

\(T = \frac{2}{{3K + 1}}\)

The time constant of the system is less than 0.2

\( \Rightarrow \frac{2}{{3K + 1}} < 0.2\;\)

⇒ K > 3   

Concept:

The time constant (τ) of an RC circuit is given by the formula:

τ = R_eqC

where R_eq is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

R_eq = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = R_eqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Concept:

The root locus of a system is the locus of closed-loop poles as k varies from 0 to ∞.  For, k = 0, the closed-loop poles are equal to the open-loop poles and for k = ∞, the closed-loop poles is equal to the open-loop zeros. 

Analysis:

When k is varied from 0 to ∞, the roots of the characteristic equation are from open-loop pole to open-loop zero.

When k = 0, the location of the pole is:

\(s = \frac{{ - 1}}{{0.01}} = - 100\)

The time constant is:

\(\tau = \frac{{ - 1}}{{Real\;part\;of\;pole}}\)

\(\tau = \frac{{ - 1}}{{ - 100}} = 0.01\)

When k = ∞, the location of the zero is:

\(\frac{{ - 1}}{{0.1}} = - 10\)

The time constant is:

\(\tau = \frac{{ - 1}}{{Real\;part\;of\;zero}}\)

\(\tau = \frac{{ - 1}}{{ - 10}} = 0.1\)