In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

Class 12 Exam > Class 12 Questions > In the given circuit if switches S3 and S4 ar...

Join the Discussion on the App

In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is?

Join for Free

Join for Free

Most Upvoted Answer

In the given circuit if switches S3 and S4 are open keeping S1And S2 c...

Introduction:
In the given circuit, there are four switches (S1, S2, S3, and S4) connected in different combinations. The behavior of the circuit depends on the combination of switches that are closed or open.

Explanation:
When switches S3 and S4 are open and S1 and S2 are closed, the value of current I is I0.

When switches S1 and S2 are open and S3 and S4 are closed, the value of current I is I0 sin wt.

Now, let's consider the scenario where all switches are closed.

I. All switches closed:
When all switches (S1, S2, S3, and S4) are closed, the circuit is complete, and current flows through it. In this scenario, the RMS current through the AC component during one complete cycle can be calculated using the following steps:

1. Identify the circuit components: The circuit consists of resistors (R), capacitors (C), and inductors (L) connected in series or parallel.

2. Calculate the impedance: For each component, calculate the impedance (Z) using the respective formulas: Z = R for resistors, Z = 1 / (jωC) for capacitors, and Z = jωL for inductors, where ω is the angular frequency.

3. Find the total impedance: Combine the impedances of all components connected in series or parallel to find the total impedance (Z_total) of the circuit.

4. Calculate the RMS current: Use Ohm's law (I = V / Z_total) to calculate the RMS current (I) flowing through the circuit, where V is the voltage across the circuit.

5. Determine the AC component: The AC component of the current can be obtained by taking the imaginary part of the complex current.

6. Calculate the RMS current: Finally, calculate the RMS value of the obtained AC component of the current.

Conclusion:
In the given circuit, if all switches are closed, the RMS current through the AC component during one complete cycle can be calculated by considering the circuit components, calculating the total impedance, and applying Ohm's law. The AC component of the current will depend on the specific values of resistors, capacitors, and inductors in the circuit.

View all answers Start learning for free

Explore Courses for Class 12 exam

Similar Class 12 Doubts

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What are the main 3 components in a Roget’s spiral?

View answer

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Why the spring shrinks in Roget’s spiral ?

View answer

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What else can be used instead of mercury in Roget’s spiral ?

View answer

Read the following text and answer the following questions on the basis of the same:Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Magnetic effects

View answer

A battery of emf V volt, resistors R1 and R2, a condenser C and switches S1 and S2 are connected in the circuit as shown in the figure below:?

View answer

Share with a Friend

Answer this doubt

Question Description
In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is?.

Solutions for In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.

Here you can find the meaning of In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? defined & explained in the simplest way possible. Besides giving the explanation of In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is?, a detailed solution for In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? has been provided alongside types of In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? theory, EduRev gives you an ample number of questions to practice In the given circuit if switches S3 and S4 are open keeping S1And S2 closed the value of I is I0 if switches S1 and S2 are open keeping S3 And S4 closed the value of I is I0 sin wt now if all switches are closed RMS current through the ac component during one complete cycle is? tests, examples and also practice Class 12 tests.

Explore Courses for Class 12 exam