![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
A circuit contains two capacitors in parallel wired in series to a resistor and source as shown above. The capacitors are initially uncharged. Which of the following describes the current I through and voltage $V_c$ across the Capacitor $C_1$ after the circuit reaches steady state?
![](https://www.vedantu.com/question-sets/e04fe6b6-196d-4da9-8a16-ab0017ac0a5d4211530711461834622.png)
A) I=0 , ${V_c}$=0
B) I=0,${V_c}$=V
C) I=0, ${V_c}$=V/2
D) I=V/R, ${V_c}$=0
E) I=V/R, ${V_c}$=V
Answer
126k+ views
Hint: Experimentally the time taken for a capacitor to get completely charged is equal to t=RC where t is the time constant in charging of the capacitor in C-R circuit, R is the resistance of the resistor in series and C is the capacitance of the capacitor. From this we can conclude that adding a resistance in series will increase the time taken by the capacitor to get fully charged.
Complete step by step answer:
A capacitor is a charge storing device. It is connected to a resistor initially and charged through a battery. The charging process is time dependent and it takes some time depending on the RC value of the circuit for the capacitor to reach maximum charge. But since we are leaving the capacitor connected for a long time, we assume that it is storing its maximum charge.
A capacitor connected across a DC supply keeps on charging till the potential across the battery becomes equal to the potential difference across the capacitor. In steady state after some time the current in the circuit will numerically be equal to zero. Hence we will use Kirchhoff’s law of voltage to see in steady state the potential difference across the capacitor.
Therefore, I=0, ${V_c}$=V and the correct option is (B).
Note: In an RC circuit, if the source is DC, the current then decreases from its initial value of I to zero as the voltage of the capacitor reaches the same value as the emf in case of transient period i.e. for a very short time in microseconds. As we know capacitors block the DC current, meaning the circuit will act as an open circuit.
Complete step by step answer:
A capacitor is a charge storing device. It is connected to a resistor initially and charged through a battery. The charging process is time dependent and it takes some time depending on the RC value of the circuit for the capacitor to reach maximum charge. But since we are leaving the capacitor connected for a long time, we assume that it is storing its maximum charge.
A capacitor connected across a DC supply keeps on charging till the potential across the battery becomes equal to the potential difference across the capacitor. In steady state after some time the current in the circuit will numerically be equal to zero. Hence we will use Kirchhoff’s law of voltage to see in steady state the potential difference across the capacitor.
Therefore, I=0, ${V_c}$=V and the correct option is (B).
Note: In an RC circuit, if the source is DC, the current then decreases from its initial value of I to zero as the voltage of the capacitor reaches the same value as the emf in case of transient period i.e. for a very short time in microseconds. As we know capacitors block the DC current, meaning the circuit will act as an open circuit.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Young's Double Slit Experiment Step by Step Derivation
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Login 2045: Step-by-Step Instructions and Details
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Derivation of Equation of Trajectory in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)