Q19.For the given circuit, in the steady state, |VB −VD| = _____ V .
What This Question Tests
This question tests the understanding of steady-state behavior of a capacitor in a DC circuit, where it acts as an open circuit, and applying Ohm's law to find potential differences.
Concepts Tested
Formulas Used
I = V/R
In steady state, capacitor acts as open circuit
V_C = Q/C
📚 NCERT Sections This Tests
8.1 — Figure 8.5 Shows A Capacitor Made Of Two Circular Plates Each Of
Physics Class 11 · Chapter 8
8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. FIGURE 8.5
3.2 — A Battery Of Emf 10 V And Internal Resistance 3 Ω Is Connected To A
Physics Class 11 · Chapter 3
3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
2.6 — Three Capacitors Each Of Capacitance 9 Pf Are Connected In Series.
Physics Class 11 · Chapter 2
2.6 Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
📋 Question Details
- Chapter
- Current Electricity
- Topic
- RC Circuits (Steady State)
- Year
- 2023
- Shift
- 31 Jan Shift 2
- Q Number
- Q19
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 3: Current Electricity
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