JEE Exam > JEE Questions > A resistor ‘R’ and 2µF capa...
Start Learning for Free
A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)
- a)1.7 × 105 Ω
- b)2.7 × 106 Ω
- c)3.3 × 107 Ω
- d)1.3 × 104 Ω
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A resistor ‘R’ and 2µF capacitor in series is connec...
We have, V = V0 (1 – e–t/RC)
⇒ 120 = 200 (1 - e-t / RC)
⇒ t = RC in (2.5)
⇒ 120 = 200 (1 - e-t / RC)
⇒ t = RC in (2.5)
⇒ R = 2.71 × 106 W
Most Upvoted Answer
A resistor ‘R’ and 2µF capacitor in series is connec...
A resistor is an electrical component that limits or controls the flow of electric current in a circuit. It is designed to have a specific resistance, which is measured in ohms (Ω). Resistors are typically made of materials with high resistance, such as carbon or metal films, and they come in various shapes and sizes.
Resistors are used in electronic circuits for a variety of purposes, including reducing current flow, dividing voltage, and controlling signal levels. They can also be used to generate heat in certain applications, such as in electric heaters or light bulbs.
The resistance value of a resistor determines how much it restricts the flow of current. A higher resistance value will result in a lower current flow, while a lower resistance value will allow more current to pass through. The relationship between current, voltage, and resistance is described by Ohm's Law, which states that the current flowing through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance.
Resistors are typically color-coded to indicate their resistance value, with a set of colored bands representing different digits and a multiplier. By interpreting the color code, one can determine the resistance value of a resistor. Additionally, resistors are often labeled with their resistance value in ohms, such as 100Ω for a resistor with a resistance of 100 ohms.
Overall, resistors play a crucial role in controlling and manipulating the flow of electric current in electronic circuits, making them essential components in various electronic devices and systems.
Resistors are used in electronic circuits for a variety of purposes, including reducing current flow, dividing voltage, and controlling signal levels. They can also be used to generate heat in certain applications, such as in electric heaters or light bulbs.
The resistance value of a resistor determines how much it restricts the flow of current. A higher resistance value will result in a lower current flow, while a lower resistance value will allow more current to pass through. The relationship between current, voltage, and resistance is described by Ohm's Law, which states that the current flowing through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance.
Resistors are typically color-coded to indicate their resistance value, with a set of colored bands representing different digits and a multiplier. By interpreting the color code, one can determine the resistance value of a resistor. Additionally, resistors are often labeled with their resistance value in ohms, such as 100Ω for a resistor with a resistance of 100 ohms.
Overall, resistors play a crucial role in controlling and manipulating the flow of electric current in electronic circuits, making them essential components in various electronic devices and systems.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Question Description
A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer?.
A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)a)1.7 × 105 Ωb)2.7 × 106 Ωc)3.3 × 107 Ωd)1.3 × 104 ΩCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily