Question

A resistor of $200\Omega $ and a capacitor of $15\mu F$ are connected in series to a $220V,50Hz$, A.C source . Calculate the current in the circuit and the arm voltage across the resistor and capacitor. Why is the algebraic sum of these voltages more than the source voltage?

Answer 1

Hint: Before going through the question we should know about the alternating current. A current that changes its magnitude and polarity at regular intervals is known as alternating current. It can also be described as an electrical current that alternates or reverses its direction on a regular basis, as opposed to Direct Current (DC), which only flows in one direction.

Complete answer:
Given,
R=$200\Omega $ ; Capacitance of capacitor$\left( c \right) = 15 \times {10^{ - 6}}F$
$
  {V_{rms}} = 220V\,, \\
  frequency\left( f \right) = 50Hz \\
 $
Impedance:
\[
  \left( Z \right) = \sqrt {{R^2} + {X_c}^2} \\
  \,\,\,\,\,\,\,\, = \sqrt {{{\left( {200} \right)}^2} + {{\left( {\dfrac{1}{{c\omega }}} \right)}^2}} \\
 \]
Relation of$\omega $with frequency $\left( f \right)$
$\omega = 2\pi f$
$\omega = \left( {2\pi } \right) \times \left( {50} \right)$
$\omega = 100\pi \,rad\,{s^{ - 1}}$
Now, Impedance:
 $
  \left( Z \right) = \sqrt {{{\left( {200} \right)}^2} + \left( {\dfrac{1}{{50 \times {{10}^{ - 6}} \times 100{\pi ^2}}}} \right)} \\
  \therefore \left( Z \right) = 291.5 \\
 $

 So, now we can find current from the circuit
${V_{rms}} = \left( Z \right) \times {I_{rms}}$
Or, ${I_{rms}} = \dfrac{{{V_{rms}}}}{{\left( Z \right)}}$
$\therefore {I_{rms}} = \dfrac{{220}}{{291.5}} = 0.755A$
Now, voltage across the resistor and capacitor are
$
  {V_R} = \left( {{I_{rms}}} \right) \times \left( R \right) \\
  {V_R} = \left( {0.755} \right) \times \left( {200\Omega } \right) = 151V \\
$
$
  Similarly, \\
  {V_C} = \left( {{I_{rms}}} \right) \times \left( {{X_C}} \right) \\
  {V_C} = \left( {0.755} \right) \times \left( {\dfrac{1}{C}} \right)\,\, \\
  \,\,\,\,\,\,\, = \left( {0.755} \right) \times \left( {\dfrac{1}{{50 \times {{10}^{ - 6}} \times 100\pi }}} \right) = 160V \\
 $
The algebraic number of the two voltages ${V_R}and\,{V_C} = $=$(151 + 160)V = 331V$
The algebraic number of the two voltages${V_R}$ and ${V_C}$ is$311V$, which is greater than the voltage at the source. As a result, if the phase difference between two voltages is properly considered, the total voltage around the resistor and inductor equals the source voltages.

Note:
A circuit's or device's admittance is a measurement of how quickly a current can pass through it. It is known as the reciprocal of impedance, in the same way that conductance and resistance are. The siemens (symbol S) is the SI unit of admission; the older, synonymous unit is the mho, which has the symbol (an upside-down uppercase omega).