(1)A \(100~\mu F\) capacitor in series with a \(40~\Omega \) resistance is connected to a \(110~V, 60~Hz\) supply.
What is the maximum current in the circuit?
(2) A \(100~\mu F\) capacitor in series with a \(40~\Omega \) resistance is connected to a \(110~V, 60~Hz\) supply.
What is the time lag between the current maximum and the voltage maximum?
What is the maximum current in the circuit?
(2) A \(100~\mu F\) capacitor in series with a \(40~\Omega \) resistance is connected to a \(110~V, 60~Hz\) supply.
What is the time lag between the current maximum and the voltage maximum?
(1) Step 1: Find circuit impedance
Given,\( V_{rms} = 110~V, f= 60~Hz\)
\(C=100~\mu F, R=40~\Omega \)
The circuit impedance is given by \( Z=\sqrt{ R^2+ X_C^2}= \sqrt{R^2+ \left ( \dfrac{1}{\omega C} \right )^2} \)
\( Z= \sqrt{(40)^2+ \left ( \dfrac{1}{2 \pi \times 60 \times 100 \times 10^{-6}} \right )^2} \)
\( Z= \sqrt{(40)^2+ (26.52 )^2} \)
\(Z= 47.99 ~\Omega \approx 48~\Omega \)
Step 2: Find maximum current
For an RC circuit if, \(V= V_0 \sin \omega t\)
In RC circuit, current \((I)\) lags from voltage \((V)\).
\(I= I_0 \sin( \omega t + \phi )\)
\(I = \dfrac{V_0}{Z} \sin( \omega t + \phi )\)
Then the maximum current is,
\(I_0 = \dfrac{V_0}{Z} \) ....(i)
Where, \( V_0 = \sqrt{2} V_{rms} = 110 \sqrt{2} V\)
Hence from equation (i),
\( I_0 = \dfrac{110 \sqrt{2} }{48} =3.24~A\)
\( I_0 =3.24~A \)
Final Answer: \( 3.24~A \)
(2)Step 1: Calculate phase angle
Given,\( V_{rms} = 110~V, f= 60~Hz\)
\(C=100~\mu F, R=40~\Omega \)
Inductive reactance,
\( X_C = \dfrac{1}{ \omega C} =\dfrac{1}{ 2 \pi f C}\)
\( = \dfrac{1}{ 2 \pi \times 60 \times 100 \times 10^{-6}}\)
\( = 26.52 \Omega \)
Now, from phasor diagram
\( \tan \phi = \dfrac{X_C}{R}\)
\( \phi = \tan^{-1} \left (\dfrac{-26.52 }{40} \right ) \)
\( \phi = -33.5^{\circ}\)
Step 2: Find the time lag between the voltage maximum and the current maximum
For an RC circuit let, \(V= V_0 \cos \omega t\)
Voltage is maximum when \(t\) is zero.
Current in a RC circuit, \( I= I_0 \cos ( \omega t+ \phi)\)
Current is maximum when
\( \omega t+ \phi =0\)
\( \Rightarrow =t = - \dfrac{ \phi }{\omega }\) ....(i)
\( t= \dfrac{ -33.5 ^{\circ} \times \dfrac{ \pi} {180^{\circ} } }{2 \pi \times 60}\)
\( t= 1.55 \times 10^{-3} s \)
Final Answer: \(1.55~m s\)
Given,\( V_{rms} = 110~V, f= 60~Hz\)
\(C=100~\mu F, R=40~\Omega \)
The circuit impedance is given by \( Z=\sqrt{ R^2+ X_C^2}= \sqrt{R^2+ \left ( \dfrac{1}{\omega C} \right )^2} \)
\( Z= \sqrt{(40)^2+ \left ( \dfrac{1}{2 \pi \times 60 \times 100 \times 10^{-6}} \right )^2} \)
\( Z= \sqrt{(40)^2+ (26.52 )^2} \)
\(Z= 47.99 ~\Omega \approx 48~\Omega \)
Step 2: Find maximum current
For an RC circuit if, \(V= V_0 \sin \omega t\)
In RC circuit, current \((I)\) lags from voltage \((V)\).
\(I= I_0 \sin( \omega t + \phi )\)
\(I = \dfrac{V_0}{Z} \sin( \omega t + \phi )\)
Then the maximum current is,
\(I_0 = \dfrac{V_0}{Z} \) ....(i)
Where, \( V_0 = \sqrt{2} V_{rms} = 110 \sqrt{2} V\)
Hence from equation (i),
\( I_0 = \dfrac{110 \sqrt{2} }{48} =3.24~A\)
\( I_0 =3.24~A \)
Final Answer: \( 3.24~A \)
(2)Step 1: Calculate phase angle
Given,\( V_{rms} = 110~V, f= 60~Hz\)
\(C=100~\mu F, R=40~\Omega \)
Inductive reactance,
\( X_C = \dfrac{1}{ \omega C} =\dfrac{1}{ 2 \pi f C}\)
\( = \dfrac{1}{ 2 \pi \times 60 \times 100 \times 10^{-6}}\)
\( = 26.52 \Omega \)
Now, from phasor diagram
\( \tan \phi = \dfrac{X_C}{R}\)
\( \phi = \tan^{-1} \left (\dfrac{-26.52 }{40} \right ) \)
\( \phi = -33.5^{\circ}\)
Step 2: Find the time lag between the voltage maximum and the current maximum
For an RC circuit let, \(V= V_0 \cos \omega t\)
Voltage is maximum when \(t\) is zero.
Current in a RC circuit, \( I= I_0 \cos ( \omega t+ \phi)\)
Current is maximum when
\( \omega t+ \phi =0\)
\( \Rightarrow =t = - \dfrac{ \phi }{\omega }\) ....(i)
\( t= \dfrac{ -33.5 ^{\circ} \times \dfrac{ \pi} {180^{\circ} } }{2 \pi \times 60}\)
\( t= 1.55 \times 10^{-3} s \)
Final Answer: \(1.55~m s\)
