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Electromagnetic Waves
Displacement Current
6 questions from 6 papers, each with its answer.
- questions
- 6
- papers
- 6
- years asked, 2017–2026
- 4 of 10
- of Electromagnetic Waves
- 12.24%
Appears in
Concepts tested
- capacitive reactance vs frequency
- conduction and displacement current for a charging capacitor
- Displacement current
- Displacement current between capacitor plates
- displacement current in a capacitor
- displacement-current-peak
All Questions
2024 1 question
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A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates;During charging, the conduction current I flows through the wires and resistor, while the changing electric field between the plates produces a displacement current $I_d = \varepsilon_0\frac{d\Phi_E}{dt}$. According to the Ampere-Maxwell law, the displacement current is equal in magnitude to the conduction current I and flows in the same direction, keeping the current continuous.
2023 1 question
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An ac source is connected to a capacitor $C$. Due to decrease in its operating frequency:
Lower frequency raises $X_C$, so displacement current decreases.
Capacitive reactance $=\frac{1}{\omega C}=X_c$ (say) On decreasing the operating frequency, $\omega$ reduces. As $X_c$ is inversely proportional to $\omega$, the value of $X_c$ increases. $\therefore I_C=I_D=\frac{V_o}{X_c}$ As $X_c$ increases, therefore displacement current $I_d$ decreases.
2023 Manipur 1 question
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To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 $\mu F$, the rate of change of applied variable potential difference $\left(\dfrac{dV}{dt}\right)$ must be:$Q=CV\Rightarrow\dfrac{dQ}{dt}=C\dfrac{dV}{dt}\Rightarrow\dfrac{dV}{dt}=\dfrac{I}{C}=\dfrac{2\times10^{-3}}{4\times10^{-6}}=\dfrac{10^3}{2}=500$ V/s.
2021 1 question
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A capacitor of capacitance $C$, is connected across an ac source of voltage $V$, given by $V=V_0\sin\omega t$. The displacement current between the plates of the capacitor, would then be given by:$q = CV \Rightarrow I_d = \frac{dq}{dt} = C\frac{dV}{dt}$ $I_d = C\frac{d}{dt}(V_0\sin\omega t) = C(V_0\omega\cos\omega t)$ $I_d = V_0\omega C\cos\omega t$
2019 1 question
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A parallel plate capacitor of capacitance $20\ \mu F$ is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:$C=20\ \mu F=20\times10^{-6}\ F$, $\frac{dV}{dt}=3\ V/s$ $q=CV\Rightarrow i_c=\frac{dq}{dt}=C\frac{dV}{dt}=20\times10^{-6}\times3=60\times10^{-6}\ A=60\ \mu A$ The displacement current equals the conduction current: $i_d=i_c=60\ \mu A$
2016 Phase II 1 question
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A 100 $\Omega$ resistance and a capacitor of 100 $\Omega$ reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is:
Peak displacement current equals peak conduction current.
$(i_d)_{max} = (i_c)_{max} = i_0 = \frac{\varepsilon_0}{Z} = \frac{220\sqrt{2}}{\sqrt{100^2 + 100^2}}$
$(i_d)_{max} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2$ A
As the amplitude of displacement current is asked, the charge on the capacitor does not matter.