Alternating Current

Concept Of Physics

H C Verma

1   Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.

Solution :

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

$f = 50Hz$ $\\$ $I = I_0 Sin Wt$ $\\$ Peak Value $I = \frac{I_0}{\sqrt{2}}$ $\\$ $\frac{I_0}{\sqrt{2}} = I_0 Sin Wt$ $\\$ $\Rightarrow \frac{1}{\sqrt{2}} = Sin Wt = Sin\frac{\pi}{4}$ $\\$ $\Rightarrow \frac{\pi}{4} = Wt, \qquad or, \quad t =\frac{\pi}{400} = \frac{\pi}{4 \times 2 \pi f} = \frac{1}{8f} = \frac{1}{8 \times 50} = 0.0025s =2.5ms$

2   The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.

Solution :

$E_{rms} = 220V$ $\\$ Frequency $= 50Hz$ $\\$ (a) $E_{rms} = \frac{E_0}{\sqrt{2}}$ $\\$ $\Rightarrow E_0 = E_{rms}\sqrt{2} = \sqrt{2} \times 220 = 1.414 \times 220 = 311.08 V = 311V$ $\\$ (b) Time taken for the current to reach the peak value = Time taken to reach the 0 value from r.m.s $\\$ $I =\frac{I_0}{\sqrt{2}} \Rightarrow \frac{I_0}{\sqrt{2}} = I_0 Sin \omega t$ $\\$ $\Rightarrow \omega t = \frac{\pi}{4}$ $\\$ $\Rightarrow t = \frac{\pi}{4 \omega } = \frac{\pi}{4 \times 2 \pi f} = \frac{\pi}{ 8 \pi 50} = \frac{1}{400} = 2.5 ms$

3   A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.

Solution :

$P = 60W, \qquad V =20V =E$ $\\$ $R = \frac{V^2}{P} = \frac{220 \times 220 }{60} = 806.67$ $\\$ $\epsilon_0 = \sqrt{2} E = 1.414 \times 220 = 311.08$ $\\$ $I_0 = \frac{\epsilon_0}{R} = \frac{806.67}{311.08} = 0.385 = 0.39 A$

4   An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source ?

Solution :

$E = 12 volts$ $\\$ $i^2 Rt = i^2_{rms} RT$ $\\$ $\Rightarrow \frac{E^2}{R^2} = \frac{E^2_{rms}}{R^2} \Rightarrow E^2 = \frac{E_0^2}{2}$ $\\$ $\Rightarrow E_0 = \sqrt{2 \times 144 } = 16.97 = 17V$

5   The peak power consumed by a resistive coil when connected to an AC source is 80 W. Find the energy consumed by the coil in 100 seconds which is many times larger than the time period of the source.

Solution :

$P_0 = 80W$ (given) $\\$ $P_{rms} = \frac{P_0}{2} = 40 W$ $\\$ Energy consumed $= P \times t = 40 \times 100 = 4000 J = 4.0 KJ$

6   The dielectric strength of air is $3.0 \times 10^6 V/m$. A parallel-plate air-capacitor has area $20 cm^ 2$ and plate separation 0.10 mm. Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Solution :

$E = 3 \times 10^6 V/m, \qquad A = 20 cm^2 , \qquad d = 0.1 mm$ $\\$ Potential diff. across the capacitor $= Ed = 3 \times 10 ^6 \times 0.1 \times 10 –3 = 300 V$ $\\$ Max. rms Voltage $= \frac{V}{\sqrt{2}} = \frac{300}{\sqrt{2}} = 212 V$

7   The current in a discharging LR circuit is given by $i - i_0 e^{-t/ \tau}$ where T is the time constant of the circuit. Calculate the rms current for the period $t = 0 \quad to \quad t = \tau.$

Solution :

$i = i_0 e^{-t/r}$ $\\$ $\vec{i^2} = \frac{1}{\tau} \int_{0}^{\tau}i_0^2e^{-2t/ \tau} dt = \frac{i_0^2}{\tau}\int_{0}^{\tau}e^{-2t/ \tau} dt = \frac{i_0^2}{\tau} \times \Bigg[ \frac{\tau}{2}e^{-2t/ \tau} \Bigg] _{0}^{\tau} = -\frac{i_0^2}{\tau} \times \frac{\tau}{2} \times \Bigg[ e^{-2} -1 \Bigg]$ $\\$ $\sqrt{\vec{i^2}} = \sqrt{-\frac{i_0^2}{2} \Bigg( \frac{1}{e^2} -1\Bigg)} = \frac{i_0}{e} \sqrt{\Bigg( \frac{e^2 - 1}{2} \Bigg)}$

8   A capacitor of capacitance $10 \mu F$ is connected to an oscillator giving an output voltage $\epsilon = (10 V)sin \omega t$. Find the peak currents in the circuit for $\omega = 10s^{-1}, 100 s^{-1} , 500 s^{-1} , 1000 s^{-1}$ .

Solution :

$C =10 \mu F = 10 \times 10^{-6} F = 10^{-5}F$ $\\$ $E = (10V) Sin \omega t$ $\\$ (a) $I =\frac{E_0}{Xc} = \frac{E_0}{\Bigg( \frac{1}{\omega C} \Bigg)} = \frac{10}{\Bigg( \frac{1}{10 \times 10^{-5}} \Bigg)} = 1 \times 10^{-3} A$ $\\$ (b) $\omega = 100 s^{-1}$ $\\$ $I = \frac{E_0}{\Bigg( \frac{1}{\omega C} \Bigg)} = \frac{10}{\Bigg( \frac{1}{100 \times 10^{-5}} \Bigg)} = 1 \times 10^{-2} A = 0.01A$ $\\$ (c) $\omega =500 s^{-1}$ $\\$ $I = \frac{E_0}{\Bigg( \frac{1}{\omega C} \Bigg)} = \frac{10}{\Bigg( \frac{1}{500 \times 10^{-5}} \Bigg)} = 5 \times 10^{-2} A = 0.05 A$ $\\$ (d) $\omega =1000 s^{-1}$ $\\$ $I = \frac{E_0}{\Bigg( \frac{1}{\omega C} \Bigg)} = \frac{10}{\Bigg( \frac{1}{1000 \times 10^{-5}} \Bigg)} = 1 \times 10^{-1} A = 0.1 A$

9   A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for $\omega =100 s^{-1} , 500 s^{-1} , 1000 s^{-1} .$

Solution :

Inductance $= 5.0mH = 0.005 H$ $\\$ (a) $\omega = 100 s^{-1}$ $\\$ $X_L = \omega L = 100 \times \frac{5}{1000} = 0.5 \Omega$ $\\$ $i = \frac{\epsilon_0}{X_L} = \frac{10}{0.5} = 20A$ $\\$ (b) $\omega = 500 s^{-1}$ $\\$ $X_L = \omega L = 500 \times \frac{5}{1000} = 2.5 \Omega$ $\\$ $i = \frac{\epsilon_0}{X_L} = \frac{10}{2.5} = 4A$ $\\$ (c) $\omega = 1000 s^{-1}$ $\\$ $X_L = \omega L = 1000 \times \frac{5}{1000} = 5 \Omega$ $\\$ $i = \frac{\epsilon_0}{X_L} = \frac{10}{5} = 2A$

10   A coil has a resistance of $10 \Omega$ and an inductance of 0.4 henry. It is connected to an AC source of $6.5 V,\frac{30}{\pi} Hz$. Find the average power consumed in the circuit.

Solution :

$R = 10 \Omega , \qquad L -0.4 Henry$ $\\$ $E = 6.5 V, \qquad f =\frac{30}{ \pi} Hz$ $\\$ $Z = \sqrt{R^2 \times X_L^2} = \sqrt{R^2 + (2 \pi f L)^2}$ $\\$ Power $= V_{rms } I_{rms} cos \phi$ $\\$ $=6.5 \times \frac{6.5}{Z} \times \frac{R}{Z} = \frac{6.5 \times 6.5 \times 10}{[ \sqrt{ R^2 +(2 \pi f L)^2}]^2 } = \frac{6.5 \times 6.5 \times 10}{10^2 + \Bigg( 2 \pi \times \frac{30}{\pi} \times 0.4 \Bigg)} = \frac{6.5 \times 6.5 \times 10}{100 + 576} = 0.625 = \frac{5}{8} \omega$

11   A resistor of resistance 100 ft is connected to an AC source $\epsilon = (12 V) sin (250 \pi s^{-1} )t$. Find the energy dissipated as heat during $t = 0 \quad to \quad t =1.0 ms.$

Solution :

$H =\frac{V^2}{R} T, \qquad E_0 = 12V, \qquad \omega =250 \pi, \qquad R =100 \Omega$ $\\$ $H = \int_{0}^{H} dH = \int \frac{E_0^2 Sin^2 \omega t}{R} dt = \frac{144}{100} \int sin^2 \omega t dt = 1.44 \int \Bigg( \frac{1 - cos2\omega t}{2} \Bigg)$ $\\$ $=\frac{1.44}{2} \Bigg[ \int_{0}^{10^{-3}} dt - \int_{0}^{10^{-3}} Cos 2 \omega t dt \Bigg] = 0.72 \Bigg[ 10^{-3} - \Bigg( \frac{Sin2 \omega t}{2 \omega} \Bigg)_{0}^{10^{-3}} \Bigg]$ $\\$ $= 0.72 \Bigg[ \frac{1}{1000} - \frac{1}{500 \pi} \Bigg] = \frac{( \pi - 2)}{1000 \pi} \times 0.72 = 0.0002614 = 2.61 \times 10^{-4} J$

12   In a series RC circuit with an AC source, $R = 300 \Omega$, $C = 25 \mu F$,$\epsilon_0 = 50 V$ and $v = \frac{50} {\pi} Hz$. Find the peak current and the average power dissipated in the circuit

Solution :

$R = 300 \Omega, \qquad C= 25 \mu F=25 \times 10^{-6}F, \qquad \epsilon_0 = 50 V, \qquad f = 50Hz$ $\\$ $X_c = \frac{1}{\omega c} = \frac{1}{\frac{50}{\pi} \times 2 \pi \times 25 \times 10^(-6) } = \frac{10^4}{25}$ $\\$ $Z= \sqrt{R^2 + X_c^2} = \sqrt{(300)^2 + \Bigg( \frac{10^4}{25} \Bigg)^2} = \sqrt{(300)^2 + (400)^2} = 500$ $\\$ (a) Peak Current $= \frac{E_0}{Z} =\frac{50}{500} = 0.1A$ $\\$ (b) Average Power dissipiated $= E_{rms}I_{rms} Cos \phi$ $= \frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2Z}} \times \frac{R}{Z} = \frac{E_0^2}{2Z^2} = \frac{50 \times 50 \times 300}{2 \times 500 \times 500} = \frac{3}{2} = 1.5 \omega$

13   An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage ?

Solution :

$Power = 55W , \qquad Voltage = 110V, \qquad Resistance = \frac{V^2}{P} = \frac{110\times 110}{55} = 220 \Omega$ $\\$ $Frequency(f) = 50 Hz, \qquad \omega = 2 \pi f = 2 \pi \times 50 = 100 \pi$ $\\$ Current in the circuit $= \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (\omega L)^2}}$ $\\$ Voltage drop across the resistor $= ir = \frac{VR}{\sqrt{R^2 + (\omega L)^2}}$ $\\$ $=\frac{220 \times 220}{\sqrt{(220)^2 +(100 \pi L)^2}} = 110$ $\\$ $\Rightarrow 220 \times 2 = \sqrt{(220)^2 +(100 \pi L)^2} \qquad \Rightarrow (220)^2 + (100 \pi L)^2 = (440)^2$ $\\$ $\Rightarrow 48400 + 10^4 \pi^2 L^2 = 193600 \qquad \Rightarrow 10^4 \pi^2 L^2 = 193600 -48400$ $\\$ $\Rightarrow L^2 = \frac{142500}{\pi^2 \times 10^4} = 1.4726 \qquad \Rightarrow L =1.2135 = 1.2 Hz$

14   In a series LCR circuit with an AC source, $R = 300 \Omega$, $C = 20 \mu F$, $L = 1.0 henry$, $\epsilon_0 = 50 V$ and $v = 50/ \pi Hz$. Find, (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Solution :

$R = 300 \Omega, \quad C = 20\mu F = 20 \times 10^{-6} F$ $\\$ $L =1 Henry, \qquad E =50V \qquad V =\frac{50}{\pi} Hz$ $\\$ (a) $I_0 = \frac{E_0}{Z},$ $\\$ $Z = \sqrt{ R^2 + (X_c - X_L)^2 } = \sqrt{(300)^2 + \Bigg( \frac{1}{2 \pi f C} - 2 \pi f L \Bigg)^2}$ $\\$ $= \sqrt{ (300)^2 + \Bigg( \frac{1}{ 2 \pi \times \frac{50}{\pi} \times 10^{-6} } - 2 \pi \times \frac{50}{\pi} \times 1 \Bigg)^2 } = \sqrt{(300)^2 + \Bigg( \frac{10^4}{20} -100 \Bigg)^2} = 500$ $\\$ $I_0 = \frac{E_0}{Z} = \frac{50}{500} = 0.1 A$ $\\$ (b) Potential across the capacitor $= i_ 0 \times X_ c = 0.1 \times 500 = 50 V$ Potential difference across the resistor $= i _0 \times R = 0.1 \times 300 = 30 V$ Potential difference across the inductor $= i _0 \times X_ L = 0.1 \times 100 = 10 V$ Rms. potential $= 50 V$ Net sum of all potential drops $= 50 V + 30 V + 10 V = 90 V$ Sum or potential drops > R.M.S potential applied.

15   Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.

Solution :

$R = 300 \Omega, \qquad C = 20 \mu F = 20 × 10^{-6} F. \qquad L = 1H, \qquad Z = 500$ (from 14) $\\$ $\epsilon = 50V, I_0 = \frac{E_0}{Z} = \frac{50}{500} = 0.1A$ $\\$ Electric Energy stored in Capacitor $= (1/2) CV^2 = (1/2) × 20 × 10^{-6} × 50 × 50 = 25 × 10^{-3} J = 25 mJ$ $\\$ Magnetic field energy stored in the coil $= (1/2) L I_ 0^2 = (1/2) × 1 × (0.1)^2 = 5 × 10^{-3} J = 5 mJ$

16   A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary ?

Solution :

Transformer works upon the principle of induction which is only possible in case of AC. $\\$ Hence when DC is supplied to it, the primary coil blocks the Current supplied to it and hence induced current supplied to it and hence induced Current in the secondary coil is zero.

17   An inductance of $2.0 H,$ a capacitance of $18 \mu F$ and a resistance of $10 k \Omega$ are connected to an AC source of $20 V$ with adjustable frequency, (a) What frequency should be chosen to maximise the current in the circuit ? (b) What is the value of this maximum current ?

Solution :

(a) For Current to be maximum in a circuit $\\$ $X_I = X_C \qquad$ (Reasonant Condition) $\\$ $\Rightarrow WL = \frac{1}{WC}$ $\\$ $\Rightarrow W^2 = \frac{1}{WC}= \frac{1}{2 \times 18 \times 10^{-6}}$ $\\$ $\Rightarrow W = \frac{10^3}{6} \Rightarrow 2 \pi f = \frac{10^3}{6}$ $\\$ $\Rightarrow f=\frac{1000}{6 \times 2 \pi} = 26.537 Hz = 27 Hz$ $\\$ (b) Maximum Current $= \frac{E}{R} \quad$(in resonance and) $\\$ $= \frac{20}{10 \times 10^3} = \frac{2}{10^3}A = 2 mA$

18   An inductor-coil, a capacitor and an AC source of rms voltage $24 V$ are connected in series. When the frequency of the source is varied, a maximum rms current of $6.0 A$ is observed. If this inductor coil is connected to a battery of emf $12 V$ and internal resistance $4.0 \Omega$, what will be the current ?

Solution :

$E_{rms} = 24 V$ $\\$ $r = 4 \Omega \qquad I_{rms}$ $\\$ $R = \frac{E}{I} = \frac{24}{6} = 4 \Omega$ $\\$ Internal Resistance $= 4 \Omega$ $\\$ Hence net resistance $= 4 + 4 = 8 \Omega$ $\\$ $\therefore$ Current $= \frac{12}{8} = 1.5 A$

19   Figure (39-E1) shows a typical circuit for low-pass filter. An AC input $V _i = 10 mV$ is applied at the left end and the output V 0 is received at the right end. Find the output voltages for $v = 10 kHz, 100kHz, 10 MHz$ and $10.0 MHz.$ Note that as the frequency is increased the output decreases and hence the name low-pass filter

Solution :

$V_1 = 10 \times 10^{-3} V$ $\\$ $R = 1 \times 10^3 \Omega$ $\\$ $C = 1 \times 10^{-9} F$ $\\$ (a) $X_c = \frac{1}{WC} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 10 \times 10^3 \times 10 \times 10^{-9}} = \frac{1}{2 \pi \times 10^{-4}} = \frac{10^4}{2 \pi} = \frac{5000}{ \pi }$ $\\$ $Z = \sqrt{R^2 + X_c^2} = \sqrt{(1 \times 10^3)^2 + \Bigg( \frac{5000}{\pi} \Bigg)^2} = \sqrt{10^6 + \Bigg( \frac{5000}{\pi} \Bigg)^2}$ $\\$ $I_0 = \frac{E_0}{Z} = \frac{V_1}{Z} = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \frac{5000}{\pi}}^2}$ $\\$ (b) $X_c = \frac{1}{WC} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 10^5 \times 10 \times 10^{-9}} = \frac{1}{2 \pi \times 10^{-3}} = \frac{10^3}{2 \pi} = \frac{500}{ \pi }$ $\\$ $Z = \sqrt{R^2 + X_c^2} = \sqrt{(10^3)^2 + \Bigg( \frac{500}{\pi} \Bigg)^2} = \sqrt{10^6 + \Bigg( \frac{500}{\pi} \Bigg)^2}$ $\\$ $I_0 = \frac{E_0}{Z} = \frac{V_1}{Z} = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \frac{500}{\pi}}^2}$ $\\$ $V_0 = I_0 X_c = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \Bigg( \frac{500}{\pi} \Bigg)^2}} \times \frac{500}{\pi} = 1.6124V = 1.6 mV$ $\\$

(c) $f = 1 MHz = 10^6 Hz$ $\\$ $X_c = \frac{1}{WC} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 10^6 \times 10 \times 10^{-9}} = \frac{1}{2 \pi \times 10^{-2}} = \frac{10^2}{2 \pi} = \frac{50}{ \pi }$ $\\$ $Z = \sqrt{R^2 + X_c^2} = \sqrt{(10^3)^2 + \Bigg( \frac{50}{\pi} \Bigg)^2} = \sqrt{10^6 + \Bigg( \frac{50}{\pi} \Bigg)^2}$ $\\$ $I_0 = \frac{E_0}{Z} = \frac{V_1}{Z} = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \frac{50}{\pi}}^2}$ $\\$ $V_0 = I_0 X_c = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \Bigg( \frac{50}{\pi} \Bigg)^2}} \times \frac{50}{\pi} = 0.16 mV$ $\\$ (d)$f = 10 MHz = 10^7 Hz$ $\\$ $X_c = \frac{1}{WC} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 10^7 \times 10 \times 10^{-9}} = \frac{1}{2 \pi \times 10^{-1}} = \frac{10}{2 \pi} = \frac{5}{ \pi }$ $\\$ $Z = \sqrt{R^2 + X_c^2} = \sqrt{(10^3)^2 + \Bigg( \frac{5}{\pi} \Bigg)^2} = \sqrt{10^6 + \Bigg( \frac{5}{\pi} \Bigg)^2}$ $\\$ $I_0 = \frac{E_0}{Z} = \frac{V_1}{Z} = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \frac{5}{\pi}}^2}$ $\\$ $V_0 = I_0 X_c = \frac{10 \times 10^{-3}}{\sqrt{10^6 + \Bigg( \frac{5}{\pi} \Bigg)^2}} \times \frac{5}{\pi} = 16 \mu V$ $\\$