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A resistor of 200 Omega and a capacitor...

A resistor of `200 Omega` and a capacitor of `15.0 mu F` are connected in series to a 220V , 50Hz ac source . (a) Calculate the current in the circuit, (b) Calculate the voltage (rms) across the resistor and the capacitor . Is the algeraic sum of these voltages more than source voltage ? if yes, resolve the paradox.

Text Solution

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` R= 200Omega, C = 15.0 muF = 15.0 xx 10^(-6) F, V= 220V, v = 50Hz`
a. In order to calculate the current, we need the impedance
`Z = sqrt(R^2 + X_C^2) = sqrt(R^2 + (2pi v C)^(-2) )`
` = sqrt( (200)^2 + (2 xx 3.14 xx 50 xx 10^(-6) )^(-2) ) = sqrt( (200)^2 + (212)^2) = 291.5 Omega`
Therefore, the current in the circuit is, `I = V/Z = (220)/(291.5) = 0.755 A`
b. Since, the current is the same throughout the circuit, we have
`V_R = IR = (0.755)(200) = 151 V`
`V_C = IX_C = (0.755) (212.3) = 160.3 V`
The algebraic sum of the two voltages, `V_R` and `V_C ` is 311.3 V which is more than the source voltage of 220V. You have learnt that, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem: `V_(R_+C) = sqrt(V_R^2 + V_C^2) = 220V`
Thus, if the phase difference between two voltage is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.
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