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प्रश्न
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 i = i1 cos [(100 π s−1)t + φ1) + i2 cos [(500π s−1)t + ϕ2]. So,
विकल्प
i1 > i2
i1 = i2
i1 < i2
The information is insufficient to find the relation between i1 and i2.
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उत्तर
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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