154989
An AC voltage source has an output of $\Delta \mathrm{V}=(200 \mathrm{~V}) \sin 2 \pi \mathrm{ft}$. This source is connected to a $100 \Omega$ resistor. RMS current in the resistance is
1 $1.41 \mathrm{~A}$
2 $2.41 \mathrm{~A}$
3 $3.41 \mathrm{~A}$
4 $0.71 \mathrm{~A}$
Explanation:
A Given, AC voltage source, $\Delta \mathrm{V}=200 \sin 2 \pi \mathrm{ft}=200 \sin \omega \mathrm{t}$ Resistance, $\mathrm{R}=100 \Omega$ Equation (i) is compare with standard equation of $\mathrm{AC}$ voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=200 \mathrm{~V}$ rms voltage of source, $\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100 \sqrt{2}$ Thus the rms current, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{R}}=\frac{100 \sqrt{2}}{100}$ $=\sqrt{2}=1.414 \mathrm{~A}$
UPSEE - 2009
Alternating Current
154990
An AC source of voltage $E=20 \sin 100 t$ is connected across a resistance $20 \Omega$. The rms value of current in the circuit is
1 $1 \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $\sqrt{2} \mathrm{~A}$
4 $2 \sqrt{2} \mathrm{~A}$
5 $\frac{1}{\sqrt{2}} \mathrm{~A}$
Explanation:
E Given, AC voltage source, $\mathrm{E}=20 \sin 100 \mathrm{t}$ Resistance, $\mathrm{R}=20 \Omega$ Standard equation of voltage, $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0} =20 \mathrm{~V}$ $\mathrm{I}_{0} =\frac{\mathrm{E}_{0}}{\mathrm{R}}$ $\mathrm{I}_{0} =\frac{20}{20}$ $\mathrm{I}_{0} =1 \mathrm{~A}$ Now, the rms current $I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~A}$
Kerala CEE- 2013
Alternating Current
154991
If $E=100 \sin (100$ t $)$ volt and $I=100 \sin$ $\left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$ are the instantaneous values of voltage and current, then the rms values of voltage and current are respectively
1 $70.7 \mathrm{~V}, 70.7 \mathrm{~mA}$
2 $70.7 \mathrm{~V}, 70.7 \mathrm{~A}$
3 $141.4 \mathrm{~V} 141.4 \mathrm{~mA}$
4 $141.4 \mathrm{~V}, 141.4 \mathrm{~A}$
5 $100 \mathrm{~V}, 100 \mathrm{~mA}$
Explanation:
A Given, Instantaneous voltage $\mathrm{E}=100 \sin (100 \mathrm{t}) \mathrm{V}$ Comparing with standard equation, $E=E_{0} \sin (\omega t) V$ $E_{0}=100 \mathrm{~V}$ $\omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of voltage, $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~V}$ Now, instantaneous current, $\mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$ Comparing with standard equation, $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of current is, $I_{\text {rms }}=\frac{I_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~mA}$
BCECE-2015
Alternating Current
154992
In a $L-R$ circuit, the value of $L$ is $\left(\frac{0.4}{\pi}\right) \mathrm{H}$ and the value of $R$ is $30 \Omega$. If in the circuit, an alternating emf of $200 \mathrm{~V}$ at 50 cycles/s is connected, the impedance of the circuit and current will be
1 $11.4 \Omega, 17.5 \mathrm{~A}$
2 $30.7 \Omega, 6.5 \mathrm{~A}$
3 $40.4 \Omega, 5 \mathrm{~A}$
4 $50 \Omega, 4 \mathrm{~A}$
5 $35 \Omega, 6.5 \mathrm{~A}$
Explanation:
D Given, Inductance $(\mathrm{L})=\left(\frac{0.4}{\pi}\right) \mathrm{H}$ Resistance $(\mathrm{R})=30 \Omega$ Voltage $(\mathrm{V})=200 \mathrm{~V}$ Frequency $(\mathrm{f})=\mathrm{Z}=50 \mathrm{cycle} / \mathrm{sec}$ Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}=2 \pi \mathrm{fL}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 50 \times \frac{0.4}{\pi}=40 \Omega$ We know that, Impedance of LR circuit is, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{(30)^{2}+(40)^{2}}=50 \Omega$ And, $\quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{200}{50}=4 \mathrm{~A}$ Hence, the impedance $(Z)=50 \Omega$ and current $(\mathrm{I})=4 \mathrm{~A}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
154989
An AC voltage source has an output of $\Delta \mathrm{V}=(200 \mathrm{~V}) \sin 2 \pi \mathrm{ft}$. This source is connected to a $100 \Omega$ resistor. RMS current in the resistance is
1 $1.41 \mathrm{~A}$
2 $2.41 \mathrm{~A}$
3 $3.41 \mathrm{~A}$
4 $0.71 \mathrm{~A}$
Explanation:
A Given, AC voltage source, $\Delta \mathrm{V}=200 \sin 2 \pi \mathrm{ft}=200 \sin \omega \mathrm{t}$ Resistance, $\mathrm{R}=100 \Omega$ Equation (i) is compare with standard equation of $\mathrm{AC}$ voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=200 \mathrm{~V}$ rms voltage of source, $\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100 \sqrt{2}$ Thus the rms current, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{R}}=\frac{100 \sqrt{2}}{100}$ $=\sqrt{2}=1.414 \mathrm{~A}$
UPSEE - 2009
Alternating Current
154990
An AC source of voltage $E=20 \sin 100 t$ is connected across a resistance $20 \Omega$. The rms value of current in the circuit is
1 $1 \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $\sqrt{2} \mathrm{~A}$
4 $2 \sqrt{2} \mathrm{~A}$
5 $\frac{1}{\sqrt{2}} \mathrm{~A}$
Explanation:
E Given, AC voltage source, $\mathrm{E}=20 \sin 100 \mathrm{t}$ Resistance, $\mathrm{R}=20 \Omega$ Standard equation of voltage, $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0} =20 \mathrm{~V}$ $\mathrm{I}_{0} =\frac{\mathrm{E}_{0}}{\mathrm{R}}$ $\mathrm{I}_{0} =\frac{20}{20}$ $\mathrm{I}_{0} =1 \mathrm{~A}$ Now, the rms current $I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~A}$
Kerala CEE- 2013
Alternating Current
154991
If $E=100 \sin (100$ t $)$ volt and $I=100 \sin$ $\left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$ are the instantaneous values of voltage and current, then the rms values of voltage and current are respectively
1 $70.7 \mathrm{~V}, 70.7 \mathrm{~mA}$
2 $70.7 \mathrm{~V}, 70.7 \mathrm{~A}$
3 $141.4 \mathrm{~V} 141.4 \mathrm{~mA}$
4 $141.4 \mathrm{~V}, 141.4 \mathrm{~A}$
5 $100 \mathrm{~V}, 100 \mathrm{~mA}$
Explanation:
A Given, Instantaneous voltage $\mathrm{E}=100 \sin (100 \mathrm{t}) \mathrm{V}$ Comparing with standard equation, $E=E_{0} \sin (\omega t) V$ $E_{0}=100 \mathrm{~V}$ $\omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of voltage, $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~V}$ Now, instantaneous current, $\mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$ Comparing with standard equation, $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of current is, $I_{\text {rms }}=\frac{I_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~mA}$
BCECE-2015
Alternating Current
154992
In a $L-R$ circuit, the value of $L$ is $\left(\frac{0.4}{\pi}\right) \mathrm{H}$ and the value of $R$ is $30 \Omega$. If in the circuit, an alternating emf of $200 \mathrm{~V}$ at 50 cycles/s is connected, the impedance of the circuit and current will be
1 $11.4 \Omega, 17.5 \mathrm{~A}$
2 $30.7 \Omega, 6.5 \mathrm{~A}$
3 $40.4 \Omega, 5 \mathrm{~A}$
4 $50 \Omega, 4 \mathrm{~A}$
5 $35 \Omega, 6.5 \mathrm{~A}$
Explanation:
D Given, Inductance $(\mathrm{L})=\left(\frac{0.4}{\pi}\right) \mathrm{H}$ Resistance $(\mathrm{R})=30 \Omega$ Voltage $(\mathrm{V})=200 \mathrm{~V}$ Frequency $(\mathrm{f})=\mathrm{Z}=50 \mathrm{cycle} / \mathrm{sec}$ Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}=2 \pi \mathrm{fL}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 50 \times \frac{0.4}{\pi}=40 \Omega$ We know that, Impedance of LR circuit is, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{(30)^{2}+(40)^{2}}=50 \Omega$ And, $\quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{200}{50}=4 \mathrm{~A}$ Hence, the impedance $(Z)=50 \Omega$ and current $(\mathrm{I})=4 \mathrm{~A}$
154989
An AC voltage source has an output of $\Delta \mathrm{V}=(200 \mathrm{~V}) \sin 2 \pi \mathrm{ft}$. This source is connected to a $100 \Omega$ resistor. RMS current in the resistance is
1 $1.41 \mathrm{~A}$
2 $2.41 \mathrm{~A}$
3 $3.41 \mathrm{~A}$
4 $0.71 \mathrm{~A}$
Explanation:
A Given, AC voltage source, $\Delta \mathrm{V}=200 \sin 2 \pi \mathrm{ft}=200 \sin \omega \mathrm{t}$ Resistance, $\mathrm{R}=100 \Omega$ Equation (i) is compare with standard equation of $\mathrm{AC}$ voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=200 \mathrm{~V}$ rms voltage of source, $\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100 \sqrt{2}$ Thus the rms current, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{R}}=\frac{100 \sqrt{2}}{100}$ $=\sqrt{2}=1.414 \mathrm{~A}$
UPSEE - 2009
Alternating Current
154990
An AC source of voltage $E=20 \sin 100 t$ is connected across a resistance $20 \Omega$. The rms value of current in the circuit is
1 $1 \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $\sqrt{2} \mathrm{~A}$
4 $2 \sqrt{2} \mathrm{~A}$
5 $\frac{1}{\sqrt{2}} \mathrm{~A}$
Explanation:
E Given, AC voltage source, $\mathrm{E}=20 \sin 100 \mathrm{t}$ Resistance, $\mathrm{R}=20 \Omega$ Standard equation of voltage, $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0} =20 \mathrm{~V}$ $\mathrm{I}_{0} =\frac{\mathrm{E}_{0}}{\mathrm{R}}$ $\mathrm{I}_{0} =\frac{20}{20}$ $\mathrm{I}_{0} =1 \mathrm{~A}$ Now, the rms current $I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~A}$
Kerala CEE- 2013
Alternating Current
154991
If $E=100 \sin (100$ t $)$ volt and $I=100 \sin$ $\left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$ are the instantaneous values of voltage and current, then the rms values of voltage and current are respectively
1 $70.7 \mathrm{~V}, 70.7 \mathrm{~mA}$
2 $70.7 \mathrm{~V}, 70.7 \mathrm{~A}$
3 $141.4 \mathrm{~V} 141.4 \mathrm{~mA}$
4 $141.4 \mathrm{~V}, 141.4 \mathrm{~A}$
5 $100 \mathrm{~V}, 100 \mathrm{~mA}$
Explanation:
A Given, Instantaneous voltage $\mathrm{E}=100 \sin (100 \mathrm{t}) \mathrm{V}$ Comparing with standard equation, $E=E_{0} \sin (\omega t) V$ $E_{0}=100 \mathrm{~V}$ $\omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of voltage, $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~V}$ Now, instantaneous current, $\mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$ Comparing with standard equation, $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of current is, $I_{\text {rms }}=\frac{I_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~mA}$
BCECE-2015
Alternating Current
154992
In a $L-R$ circuit, the value of $L$ is $\left(\frac{0.4}{\pi}\right) \mathrm{H}$ and the value of $R$ is $30 \Omega$. If in the circuit, an alternating emf of $200 \mathrm{~V}$ at 50 cycles/s is connected, the impedance of the circuit and current will be
1 $11.4 \Omega, 17.5 \mathrm{~A}$
2 $30.7 \Omega, 6.5 \mathrm{~A}$
3 $40.4 \Omega, 5 \mathrm{~A}$
4 $50 \Omega, 4 \mathrm{~A}$
5 $35 \Omega, 6.5 \mathrm{~A}$
Explanation:
D Given, Inductance $(\mathrm{L})=\left(\frac{0.4}{\pi}\right) \mathrm{H}$ Resistance $(\mathrm{R})=30 \Omega$ Voltage $(\mathrm{V})=200 \mathrm{~V}$ Frequency $(\mathrm{f})=\mathrm{Z}=50 \mathrm{cycle} / \mathrm{sec}$ Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}=2 \pi \mathrm{fL}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 50 \times \frac{0.4}{\pi}=40 \Omega$ We know that, Impedance of LR circuit is, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{(30)^{2}+(40)^{2}}=50 \Omega$ And, $\quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{200}{50}=4 \mathrm{~A}$ Hence, the impedance $(Z)=50 \Omega$ and current $(\mathrm{I})=4 \mathrm{~A}$
154989
An AC voltage source has an output of $\Delta \mathrm{V}=(200 \mathrm{~V}) \sin 2 \pi \mathrm{ft}$. This source is connected to a $100 \Omega$ resistor. RMS current in the resistance is
1 $1.41 \mathrm{~A}$
2 $2.41 \mathrm{~A}$
3 $3.41 \mathrm{~A}$
4 $0.71 \mathrm{~A}$
Explanation:
A Given, AC voltage source, $\Delta \mathrm{V}=200 \sin 2 \pi \mathrm{ft}=200 \sin \omega \mathrm{t}$ Resistance, $\mathrm{R}=100 \Omega$ Equation (i) is compare with standard equation of $\mathrm{AC}$ voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=200 \mathrm{~V}$ rms voltage of source, $\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100 \sqrt{2}$ Thus the rms current, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{R}}=\frac{100 \sqrt{2}}{100}$ $=\sqrt{2}=1.414 \mathrm{~A}$
UPSEE - 2009
Alternating Current
154990
An AC source of voltage $E=20 \sin 100 t$ is connected across a resistance $20 \Omega$. The rms value of current in the circuit is
1 $1 \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $\sqrt{2} \mathrm{~A}$
4 $2 \sqrt{2} \mathrm{~A}$
5 $\frac{1}{\sqrt{2}} \mathrm{~A}$
Explanation:
E Given, AC voltage source, $\mathrm{E}=20 \sin 100 \mathrm{t}$ Resistance, $\mathrm{R}=20 \Omega$ Standard equation of voltage, $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0} =20 \mathrm{~V}$ $\mathrm{I}_{0} =\frac{\mathrm{E}_{0}}{\mathrm{R}}$ $\mathrm{I}_{0} =\frac{20}{20}$ $\mathrm{I}_{0} =1 \mathrm{~A}$ Now, the rms current $I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \mathrm{~A}$
Kerala CEE- 2013
Alternating Current
154991
If $E=100 \sin (100$ t $)$ volt and $I=100 \sin$ $\left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$ are the instantaneous values of voltage and current, then the rms values of voltage and current are respectively
1 $70.7 \mathrm{~V}, 70.7 \mathrm{~mA}$
2 $70.7 \mathrm{~V}, 70.7 \mathrm{~A}$
3 $141.4 \mathrm{~V} 141.4 \mathrm{~mA}$
4 $141.4 \mathrm{~V}, 141.4 \mathrm{~A}$
5 $100 \mathrm{~V}, 100 \mathrm{~mA}$
Explanation:
A Given, Instantaneous voltage $\mathrm{E}=100 \sin (100 \mathrm{t}) \mathrm{V}$ Comparing with standard equation, $E=E_{0} \sin (\omega t) V$ $E_{0}=100 \mathrm{~V}$ $\omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of voltage, $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~V}$ Now, instantaneous current, $\mathrm{I}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$ Comparing with standard equation, $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \omega=100 \mathrm{rad} / \mathrm{s}$ The rms value of current is, $I_{\text {rms }}=\frac{I_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=70.7 \mathrm{~mA}$
BCECE-2015
Alternating Current
154992
In a $L-R$ circuit, the value of $L$ is $\left(\frac{0.4}{\pi}\right) \mathrm{H}$ and the value of $R$ is $30 \Omega$. If in the circuit, an alternating emf of $200 \mathrm{~V}$ at 50 cycles/s is connected, the impedance of the circuit and current will be
1 $11.4 \Omega, 17.5 \mathrm{~A}$
2 $30.7 \Omega, 6.5 \mathrm{~A}$
3 $40.4 \Omega, 5 \mathrm{~A}$
4 $50 \Omega, 4 \mathrm{~A}$
5 $35 \Omega, 6.5 \mathrm{~A}$
Explanation:
D Given, Inductance $(\mathrm{L})=\left(\frac{0.4}{\pi}\right) \mathrm{H}$ Resistance $(\mathrm{R})=30 \Omega$ Voltage $(\mathrm{V})=200 \mathrm{~V}$ Frequency $(\mathrm{f})=\mathrm{Z}=50 \mathrm{cycle} / \mathrm{sec}$ Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}=2 \pi \mathrm{fL}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 50 \times \frac{0.4}{\pi}=40 \Omega$ We know that, Impedance of LR circuit is, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{(30)^{2}+(40)^{2}}=50 \Omega$ And, $\quad \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{200}{50}=4 \mathrm{~A}$ Hence, the impedance $(Z)=50 \Omega$ and current $(\mathrm{I})=4 \mathrm{~A}$
