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Karnataka Board PUCPUC Science Class 11

An Ac Source Producing Emf ε = ε0 Cos (100 π S−1)T + Cos (500 π S−1)T is Connected in Series with a Capacitor and a Resistor. - Physics

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Question

An AC source producing emf ε = ε0 [cos (100 π s−1)t + cos (500 π s−1)t] is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i1 cos [(100 π s−1)t + φ1) + i2 cos [(500π s−1)t + ϕ2]. So,

Options

  • i1 > i2

  • i1 = i2

  •  i1 < i2

  • The information is insufficient to find the relation between i1 and i2.

MCQ

Solution

i1 < i2

The charge on the capacitor during steady state is given by,
`Q = C epsilon = epsilon _0C [cos(100pis^-1)]`
the steady state current is, thus, given by,
`i =(dQ)/dt = epsilon _0Cxx100pi [sin(100pis^-1)] + epsilon_0Cxx 500pi [sin({00pis)]`
c`
⇒ i = 100Cepsilon cos  [(100pis^-1)] t +  Ø_1+500piepsilon_0 cos [(500pis^-1)] + Ø_2]`
⇒ `i_1 = 100Cpiepsilon_0 & i_2 = 500C_piepsilon_0`
∴ i_2 > i_1

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Different Types of AC Circuits: AC Voltage Applied to a Capacitor
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Chapter 17: Alternating Current - MCQ [Page 329]

APPEARS IN

HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 17 Alternating Current
MCQ | Q 2 | Page 329

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