05. Dynamo, Transformer Inductance
Alternating Current

155376 How much current is given drawn by the primary coil of a transformer, which steps down $220 \mathrm{~V}$ to $55 \mathrm{~V}$ to operate a device with an impedance of $275 \Omega$ ?

1 $0.05 \mathrm{~A}$
2 $0.02 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.15 \mathrm{~A}$
Alternating Current

155379 The peak current through a $200 \mathrm{mH}$ inductor connected to an $\mathrm{AC}$ source of peak value $210 \mathrm{~V}$ and frequency $50 \mathrm{~Hz}$ is

1 $1.5 \mathrm{~A}$
2 $3.3 \mathrm{~A}$
3 $8.5 \mathrm{~A}$
4 $5.5 \mathrm{~A}$
Alternating Current

155380 Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=\mathbf{5 0} \mathrm{Hz}$ )

1 $100 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $300 \mathrm{~V}$
4 $200 \mathrm{~V}$
Alternating Current

155381 A step down transformer has a turns ratio $20 : 1$ If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, then the primary current will be

1 $4 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Alternating Current

155376 How much current is given drawn by the primary coil of a transformer, which steps down $220 \mathrm{~V}$ to $55 \mathrm{~V}$ to operate a device with an impedance of $275 \Omega$ ?

1 $0.05 \mathrm{~A}$
2 $0.02 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.15 \mathrm{~A}$
Alternating Current

155379 The peak current through a $200 \mathrm{mH}$ inductor connected to an $\mathrm{AC}$ source of peak value $210 \mathrm{~V}$ and frequency $50 \mathrm{~Hz}$ is

1 $1.5 \mathrm{~A}$
2 $3.3 \mathrm{~A}$
3 $8.5 \mathrm{~A}$
4 $5.5 \mathrm{~A}$
Alternating Current

155380 Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=\mathbf{5 0} \mathrm{Hz}$ )

1 $100 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $300 \mathrm{~V}$
4 $200 \mathrm{~V}$
Alternating Current

155381 A step down transformer has a turns ratio $20 : 1$ If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, then the primary current will be

1 $4 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Alternating Current

155376 How much current is given drawn by the primary coil of a transformer, which steps down $220 \mathrm{~V}$ to $55 \mathrm{~V}$ to operate a device with an impedance of $275 \Omega$ ?

1 $0.05 \mathrm{~A}$
2 $0.02 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.15 \mathrm{~A}$
Alternating Current

155379 The peak current through a $200 \mathrm{mH}$ inductor connected to an $\mathrm{AC}$ source of peak value $210 \mathrm{~V}$ and frequency $50 \mathrm{~Hz}$ is

1 $1.5 \mathrm{~A}$
2 $3.3 \mathrm{~A}$
3 $8.5 \mathrm{~A}$
4 $5.5 \mathrm{~A}$
Alternating Current

155380 Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=\mathbf{5 0} \mathrm{Hz}$ )

1 $100 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $300 \mathrm{~V}$
4 $200 \mathrm{~V}$
Alternating Current

155381 A step down transformer has a turns ratio $20 : 1$ If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, then the primary current will be

1 $4 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Alternating Current

155376 How much current is given drawn by the primary coil of a transformer, which steps down $220 \mathrm{~V}$ to $55 \mathrm{~V}$ to operate a device with an impedance of $275 \Omega$ ?

1 $0.05 \mathrm{~A}$
2 $0.02 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.15 \mathrm{~A}$
Alternating Current

155379 The peak current through a $200 \mathrm{mH}$ inductor connected to an $\mathrm{AC}$ source of peak value $210 \mathrm{~V}$ and frequency $50 \mathrm{~Hz}$ is

1 $1.5 \mathrm{~A}$
2 $3.3 \mathrm{~A}$
3 $8.5 \mathrm{~A}$
4 $5.5 \mathrm{~A}$
Alternating Current

155380 Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=\mathbf{5 0} \mathrm{Hz}$ )

1 $100 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $300 \mathrm{~V}$
4 $200 \mathrm{~V}$
Alternating Current

155381 A step down transformer has a turns ratio $20 : 1$ If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, then the primary current will be

1 $4 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $0.5 \mathrm{~A}$