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Electrostatic Potential And Capacitance

Question
CBSEENPH12037966

A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lav between current maximum and voltage maximum?

Solution

Here given,
Capacitance, C = 100 μF = 100 × 10-6F                      = 10-4F 
Resistance, R = 40 Ω 
 
Rms voltage, Ev = 110 voltPeak voltage, E0 = 2. Ev = 2 × 110 V Frequency of Ac supply, v = 60 Hz.,   ω = 2πv = 120π rad/sPeak current, I0 = ? 

a.) In RC circuit,  as
Z = R2+Xc2 = R2+1ω2C2
Therefore,

        I0 = E0R2+1ω2C2    = 2 × 1101600+1120 π × 10-42
       I0 = 3.24 amp. 

b.) In RC circuit, voltage lags behind the current by phase angle ϕ,
where ϕ is given by,

tan ϕ = 1/ωCR          = 1ωCR           = 1120 π × 10-4 × 40           = 0.6628 

 ϕ = tan-1(0.6628) = 33.5 °           = 33.5 π180rad. 
Hence, 

Time lag = ϕω                 = 33.5 π180 ×120 π                 = 1.55 × 10-3 sec.

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