Question

A $20\,Ω$ resistor, $1.5\,H$ inductor and $35\,μF$ capacitors are connected in series with $200\,V, 50\,Hz$ ac supply. Calculate the impedance of the circuit and also find the current through the circuit.

Answer 1

Hint: to solve this question we will first find impedance by using a formula that relates impedance, resistance, inductive reactance and capacitive reactance. After finding the value of impedance we will find current through the circuit, by dividing potential by impedance.

Formula used:
\[Z = {\sqrt {{R^2} + {{\left( {X_L - X_C} \right)}^2}} ^{}}\]
Where $Z$=impedance, $R$=resistor, $X_L$=inductive reactance , $X_C$ =capacitive reactance.
\[I = \dfrac{V}{Z}\]
Where, $I$=current in the circuit, $V$ is potential and $Z$= impedance.

Complete step by step answer:
Electrical impedance is the measurement of a circuit's resistance to current when a voltage is applied in electrical engineering. Let us look at all the given terms:
\[R{\text{ }} = {\text{ }}20\Omega ,\]
\[\Rightarrow L{\text{ }} = {\text{ }}1.5{\text{ }}H\]]
\[\Rightarrow C{\text{ }} = {\text{ }}35{\text{ }} \times {\text{ }}{10^{ - 6}}F,\;\]
\[\Rightarrow V = 220V\]
\[\Rightarrow v{\text{ }} = {\text{ }}50Hz\],
\[\Rightarrow Z{\text{ }} = {\text{ }}?\]
And \[I{\text{ }} = {\text{ }}?\].
We will substitute these values in the formula for impedance to find the value of impedance.
\[Z = {\sqrt {{R^2} + {{\left( {X_L - X_C} \right)}^2}} ^{}}\]
\[\Rightarrow Z= {\text{ }}2\pi vL{\text{ }} = {\text{ }}471{\text{ }}\Omega \]
\[\Rightarrow Z = \sqrt {{{20}^2} + {{\left( {X_L - X_C} \right)}^2}} \]

Now we need to find the values of \[X_L\] and \[X_C\]. Lets first find inductive reactance:
\[X_L{\text{ }} = {\text{ }}\omega L{\text{ }}\]
\[\Rightarrow X_L{\text{ }} = {\text{ }}2\pi vL{\text{ }}\]
\[\Rightarrow X_L{\text{ }} = {\text{ }}2 \times 3.14 \times 50 \times 1.5{\text{ }}\]
\[\Rightarrow X_L{\text{ }} = {\text{ 471}}\Omega \]
Now let's find capacitive reactance:
\[X_C{\text{ }} = {\text{ }}\dfrac{1}{{\omega C}}\]
\[\Rightarrow X_C{\text{ }} = {\text{ }}\dfrac{1}{{2\pi vC}}\]
\[\Rightarrow X_C{\text{ }} = {\text{ }}\dfrac{1}{{2. \times 3.14 \times 50 \times 35 \times {{10}^{ - 6}}}}\]
\[\Rightarrow X_C{\text{ }} = {\text{ }}\dfrac{{{{10}^6}}}{{2 \times 3.14 \times 50 \times 35}}\]
\[\Rightarrow X_C{\text{ }} = {\text{ }}\dfrac{{{{10}^6}}}{{10,990}}\]
\[\Rightarrow X_C{\text{ }} = {\text{ 90}}{\text{.99}}\,\Omega .....(2)\]
Now we will find impedance:
\[Z = \sqrt {{{20}^2} + {{\left( {471 - 90.99} \right)}^2}} \]
\[\Rightarrow Z = \sqrt {{{20}^2} + {{\left( {380} \right)}^2}} \]
\[\Rightarrow Z = \sqrt {400 + 144400} \]
\[\Rightarrow Z = \sqrt {144800} \]
\[\Rightarrow Z = 380.52\,\Omega \]
Hence the impedance is 380.52 ohms.Now let's find current through the circuit:
\[I = \dfrac{{220}}{{380.52}}\]
\[\therefore I = 0.578\,A\]

Hence the current through the circuit is \[I = 0.578\,A\].

Note:keep in mind that the connection is in series, hence only at such conditions we can apply the above mentioned formula for impedance. If the connections are in parallel, there is a different method to solve by finding individual impedance, and then adding them up. Usually we are In the habit of finding current by dividing potential by resistance, but here remember to divide potential by impedance(Z) and not resistance(R), because here impedance acts as net resistance for the circuit.