155389
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the ratio of output to input power is 0.8 then, the current drawn by primary winding is
1 $2.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $3.6 \mathrm{~A}$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{S}}=440 \mathrm{~V}$ $\mathrm{I}_{\mathrm{S}}=2.0 \mathrm{~A}$ $\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{I}}}=0.8$ We know that, Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{P}_{\mathrm{O}}}{\mathrm{P}_{\mathrm{I}}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}$ $0.8=\frac{440 \times 2.0}{220 \times \mathrm{I}_{\mathrm{P}}}$ $\mathrm{I}_{\mathrm{P}}=\frac{2 \times 2.0}{0.8}$ $\mathrm{I}_{\mathrm{P}}=5.0 \mathrm{~A}$
MHT-CET 2019
Alternating Current
155390
The input voltage of an ideal transformer is $1000 \mathrm{~V}$ and input current is $50 \mathrm{~A}$. If the output voltage is $220 \mathrm{~V}$, then find the resistance of secondary coil.
1 $2 \Omega$
2 $3 \Omega$
3 $1 \Omega$
4 $4 \Omega$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=1000 \mathrm{~V}, \mathrm{I}_{\mathrm{P}}=50 \mathrm{~A}, \mathrm{~V}_{\mathrm{S}}=220 \mathrm{~V}$ We know that, for an ideal transformer- $\mathrm{P}_{\text {in }} =\mathrm{P}_{\text {out }}$ $\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} =\frac{\mathrm{V}_{\mathrm{s}}^{2}}{\mathrm{R}_{\mathrm{s}}}$ Putting these value, we get- $1000 \times 50=\frac{(220)^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{484}{500}=.968 \approx 1 \Omega$
AIIMS-26.05.2019(E) Shift-2
Alternating Current
155391
A transformer with turns ratio $\frac{N_{1}}{N_{2}}=\frac{50}{1}$ is connected to a 120 volt $\mathrm{AC}$ supply. If primary and secondary circuit resistance are $1.5 \mathrm{k} \Omega$ and $1 \Omega$ respectively then find out power output.
1 $5.76 \mathrm{~W}$
2 $11.4 \mathrm{~W}$
3 $2.89 \mathrm{~W}$
4 $7.56 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=1.5 \mathrm{~K} \Omega$ $\mathrm{R}_{2}=1 \Omega$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{50}{1}$ $\mathrm{V}_{1}=120$ Volts We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Putting these value, we get- $\frac{50}{1}=\frac{120}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{12}{5} \mathrm{Volt}$ $\text { And } \quad \mathrm{P}_{\text {output }}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}}=\frac{\left(\frac{12}{5}\right)^{2}}{1}=\frac{144}{25}=5.76 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Alternating Current
155392
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 \mathrm{~A}$, then that in the secondary is.
1 $4 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{\mathrm{P}}\right)=140$ Number of turns secondary coil $\left(\mathrm{N}_{\mathrm{S}}\right)=280$ Primary current $\left(\mathrm{I}_{\mathrm{P}}\right)=4 \mathrm{~A}$ We know that, Transformer ratio, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{N_{P}}{N_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{140}{280}=\frac{I_{s}}{4}$ $I_{S}=2 \mathrm{~A}$
AIIMS-25.05.2019(E) Shift-2
Alternating Current
155393
A transformer consists of 500 turn in primary coil and $\mathbf{1 0}$ turns in secondary coil with the load of $10 \Omega$. Find out current in the primary coil when the voltage across secondary coil is $50 \mathrm{~V}$.
1 $5 \mathrm{~A}$
2 $\frac{1}{10} \mathrm{~A}$
3 $10 \mathrm{~A}$
4 $\frac{1}{20} \mathrm{~A}$
Explanation:
B Given, $\mathrm{N}_{1}=500$ turn $\mathrm{N}_{2}=10$ turn $\mathrm{V}_{2}=50$ volt We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Or $\quad \mathrm{V}_{1}=50 \times 50=2500 \mathrm{~V}$ And $\quad \mathrm{I}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{50}{10}=5 \mathrm{~A}$ $\therefore \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{5 \times 50}{2500}=\frac{1}{10}$ $\mathrm{I}_{1}=\frac{1}{10} \mathrm{~A}$
155389
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the ratio of output to input power is 0.8 then, the current drawn by primary winding is
1 $2.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $3.6 \mathrm{~A}$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{S}}=440 \mathrm{~V}$ $\mathrm{I}_{\mathrm{S}}=2.0 \mathrm{~A}$ $\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{I}}}=0.8$ We know that, Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{P}_{\mathrm{O}}}{\mathrm{P}_{\mathrm{I}}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}$ $0.8=\frac{440 \times 2.0}{220 \times \mathrm{I}_{\mathrm{P}}}$ $\mathrm{I}_{\mathrm{P}}=\frac{2 \times 2.0}{0.8}$ $\mathrm{I}_{\mathrm{P}}=5.0 \mathrm{~A}$
MHT-CET 2019
Alternating Current
155390
The input voltage of an ideal transformer is $1000 \mathrm{~V}$ and input current is $50 \mathrm{~A}$. If the output voltage is $220 \mathrm{~V}$, then find the resistance of secondary coil.
1 $2 \Omega$
2 $3 \Omega$
3 $1 \Omega$
4 $4 \Omega$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=1000 \mathrm{~V}, \mathrm{I}_{\mathrm{P}}=50 \mathrm{~A}, \mathrm{~V}_{\mathrm{S}}=220 \mathrm{~V}$ We know that, for an ideal transformer- $\mathrm{P}_{\text {in }} =\mathrm{P}_{\text {out }}$ $\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} =\frac{\mathrm{V}_{\mathrm{s}}^{2}}{\mathrm{R}_{\mathrm{s}}}$ Putting these value, we get- $1000 \times 50=\frac{(220)^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{484}{500}=.968 \approx 1 \Omega$
AIIMS-26.05.2019(E) Shift-2
Alternating Current
155391
A transformer with turns ratio $\frac{N_{1}}{N_{2}}=\frac{50}{1}$ is connected to a 120 volt $\mathrm{AC}$ supply. If primary and secondary circuit resistance are $1.5 \mathrm{k} \Omega$ and $1 \Omega$ respectively then find out power output.
1 $5.76 \mathrm{~W}$
2 $11.4 \mathrm{~W}$
3 $2.89 \mathrm{~W}$
4 $7.56 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=1.5 \mathrm{~K} \Omega$ $\mathrm{R}_{2}=1 \Omega$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{50}{1}$ $\mathrm{V}_{1}=120$ Volts We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Putting these value, we get- $\frac{50}{1}=\frac{120}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{12}{5} \mathrm{Volt}$ $\text { And } \quad \mathrm{P}_{\text {output }}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}}=\frac{\left(\frac{12}{5}\right)^{2}}{1}=\frac{144}{25}=5.76 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Alternating Current
155392
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 \mathrm{~A}$, then that in the secondary is.
1 $4 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{\mathrm{P}}\right)=140$ Number of turns secondary coil $\left(\mathrm{N}_{\mathrm{S}}\right)=280$ Primary current $\left(\mathrm{I}_{\mathrm{P}}\right)=4 \mathrm{~A}$ We know that, Transformer ratio, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{N_{P}}{N_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{140}{280}=\frac{I_{s}}{4}$ $I_{S}=2 \mathrm{~A}$
AIIMS-25.05.2019(E) Shift-2
Alternating Current
155393
A transformer consists of 500 turn in primary coil and $\mathbf{1 0}$ turns in secondary coil with the load of $10 \Omega$. Find out current in the primary coil when the voltage across secondary coil is $50 \mathrm{~V}$.
1 $5 \mathrm{~A}$
2 $\frac{1}{10} \mathrm{~A}$
3 $10 \mathrm{~A}$
4 $\frac{1}{20} \mathrm{~A}$
Explanation:
B Given, $\mathrm{N}_{1}=500$ turn $\mathrm{N}_{2}=10$ turn $\mathrm{V}_{2}=50$ volt We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Or $\quad \mathrm{V}_{1}=50 \times 50=2500 \mathrm{~V}$ And $\quad \mathrm{I}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{50}{10}=5 \mathrm{~A}$ $\therefore \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{5 \times 50}{2500}=\frac{1}{10}$ $\mathrm{I}_{1}=\frac{1}{10} \mathrm{~A}$
155389
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the ratio of output to input power is 0.8 then, the current drawn by primary winding is
1 $2.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $3.6 \mathrm{~A}$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{S}}=440 \mathrm{~V}$ $\mathrm{I}_{\mathrm{S}}=2.0 \mathrm{~A}$ $\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{I}}}=0.8$ We know that, Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{P}_{\mathrm{O}}}{\mathrm{P}_{\mathrm{I}}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}$ $0.8=\frac{440 \times 2.0}{220 \times \mathrm{I}_{\mathrm{P}}}$ $\mathrm{I}_{\mathrm{P}}=\frac{2 \times 2.0}{0.8}$ $\mathrm{I}_{\mathrm{P}}=5.0 \mathrm{~A}$
MHT-CET 2019
Alternating Current
155390
The input voltage of an ideal transformer is $1000 \mathrm{~V}$ and input current is $50 \mathrm{~A}$. If the output voltage is $220 \mathrm{~V}$, then find the resistance of secondary coil.
1 $2 \Omega$
2 $3 \Omega$
3 $1 \Omega$
4 $4 \Omega$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=1000 \mathrm{~V}, \mathrm{I}_{\mathrm{P}}=50 \mathrm{~A}, \mathrm{~V}_{\mathrm{S}}=220 \mathrm{~V}$ We know that, for an ideal transformer- $\mathrm{P}_{\text {in }} =\mathrm{P}_{\text {out }}$ $\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} =\frac{\mathrm{V}_{\mathrm{s}}^{2}}{\mathrm{R}_{\mathrm{s}}}$ Putting these value, we get- $1000 \times 50=\frac{(220)^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{484}{500}=.968 \approx 1 \Omega$
AIIMS-26.05.2019(E) Shift-2
Alternating Current
155391
A transformer with turns ratio $\frac{N_{1}}{N_{2}}=\frac{50}{1}$ is connected to a 120 volt $\mathrm{AC}$ supply. If primary and secondary circuit resistance are $1.5 \mathrm{k} \Omega$ and $1 \Omega$ respectively then find out power output.
1 $5.76 \mathrm{~W}$
2 $11.4 \mathrm{~W}$
3 $2.89 \mathrm{~W}$
4 $7.56 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=1.5 \mathrm{~K} \Omega$ $\mathrm{R}_{2}=1 \Omega$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{50}{1}$ $\mathrm{V}_{1}=120$ Volts We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Putting these value, we get- $\frac{50}{1}=\frac{120}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{12}{5} \mathrm{Volt}$ $\text { And } \quad \mathrm{P}_{\text {output }}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}}=\frac{\left(\frac{12}{5}\right)^{2}}{1}=\frac{144}{25}=5.76 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Alternating Current
155392
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 \mathrm{~A}$, then that in the secondary is.
1 $4 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{\mathrm{P}}\right)=140$ Number of turns secondary coil $\left(\mathrm{N}_{\mathrm{S}}\right)=280$ Primary current $\left(\mathrm{I}_{\mathrm{P}}\right)=4 \mathrm{~A}$ We know that, Transformer ratio, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{N_{P}}{N_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{140}{280}=\frac{I_{s}}{4}$ $I_{S}=2 \mathrm{~A}$
AIIMS-25.05.2019(E) Shift-2
Alternating Current
155393
A transformer consists of 500 turn in primary coil and $\mathbf{1 0}$ turns in secondary coil with the load of $10 \Omega$. Find out current in the primary coil when the voltage across secondary coil is $50 \mathrm{~V}$.
1 $5 \mathrm{~A}$
2 $\frac{1}{10} \mathrm{~A}$
3 $10 \mathrm{~A}$
4 $\frac{1}{20} \mathrm{~A}$
Explanation:
B Given, $\mathrm{N}_{1}=500$ turn $\mathrm{N}_{2}=10$ turn $\mathrm{V}_{2}=50$ volt We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Or $\quad \mathrm{V}_{1}=50 \times 50=2500 \mathrm{~V}$ And $\quad \mathrm{I}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{50}{10}=5 \mathrm{~A}$ $\therefore \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{5 \times 50}{2500}=\frac{1}{10}$ $\mathrm{I}_{1}=\frac{1}{10} \mathrm{~A}$
155389
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the ratio of output to input power is 0.8 then, the current drawn by primary winding is
1 $2.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $3.6 \mathrm{~A}$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{S}}=440 \mathrm{~V}$ $\mathrm{I}_{\mathrm{S}}=2.0 \mathrm{~A}$ $\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{I}}}=0.8$ We know that, Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{P}_{\mathrm{O}}}{\mathrm{P}_{\mathrm{I}}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}$ $0.8=\frac{440 \times 2.0}{220 \times \mathrm{I}_{\mathrm{P}}}$ $\mathrm{I}_{\mathrm{P}}=\frac{2 \times 2.0}{0.8}$ $\mathrm{I}_{\mathrm{P}}=5.0 \mathrm{~A}$
MHT-CET 2019
Alternating Current
155390
The input voltage of an ideal transformer is $1000 \mathrm{~V}$ and input current is $50 \mathrm{~A}$. If the output voltage is $220 \mathrm{~V}$, then find the resistance of secondary coil.
1 $2 \Omega$
2 $3 \Omega$
3 $1 \Omega$
4 $4 \Omega$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=1000 \mathrm{~V}, \mathrm{I}_{\mathrm{P}}=50 \mathrm{~A}, \mathrm{~V}_{\mathrm{S}}=220 \mathrm{~V}$ We know that, for an ideal transformer- $\mathrm{P}_{\text {in }} =\mathrm{P}_{\text {out }}$ $\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} =\frac{\mathrm{V}_{\mathrm{s}}^{2}}{\mathrm{R}_{\mathrm{s}}}$ Putting these value, we get- $1000 \times 50=\frac{(220)^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{484}{500}=.968 \approx 1 \Omega$
AIIMS-26.05.2019(E) Shift-2
Alternating Current
155391
A transformer with turns ratio $\frac{N_{1}}{N_{2}}=\frac{50}{1}$ is connected to a 120 volt $\mathrm{AC}$ supply. If primary and secondary circuit resistance are $1.5 \mathrm{k} \Omega$ and $1 \Omega$ respectively then find out power output.
1 $5.76 \mathrm{~W}$
2 $11.4 \mathrm{~W}$
3 $2.89 \mathrm{~W}$
4 $7.56 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=1.5 \mathrm{~K} \Omega$ $\mathrm{R}_{2}=1 \Omega$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{50}{1}$ $\mathrm{V}_{1}=120$ Volts We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Putting these value, we get- $\frac{50}{1}=\frac{120}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{12}{5} \mathrm{Volt}$ $\text { And } \quad \mathrm{P}_{\text {output }}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}}=\frac{\left(\frac{12}{5}\right)^{2}}{1}=\frac{144}{25}=5.76 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Alternating Current
155392
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 \mathrm{~A}$, then that in the secondary is.
1 $4 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{\mathrm{P}}\right)=140$ Number of turns secondary coil $\left(\mathrm{N}_{\mathrm{S}}\right)=280$ Primary current $\left(\mathrm{I}_{\mathrm{P}}\right)=4 \mathrm{~A}$ We know that, Transformer ratio, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{N_{P}}{N_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{140}{280}=\frac{I_{s}}{4}$ $I_{S}=2 \mathrm{~A}$
AIIMS-25.05.2019(E) Shift-2
Alternating Current
155393
A transformer consists of 500 turn in primary coil and $\mathbf{1 0}$ turns in secondary coil with the load of $10 \Omega$. Find out current in the primary coil when the voltage across secondary coil is $50 \mathrm{~V}$.
1 $5 \mathrm{~A}$
2 $\frac{1}{10} \mathrm{~A}$
3 $10 \mathrm{~A}$
4 $\frac{1}{20} \mathrm{~A}$
Explanation:
B Given, $\mathrm{N}_{1}=500$ turn $\mathrm{N}_{2}=10$ turn $\mathrm{V}_{2}=50$ volt We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Or $\quad \mathrm{V}_{1}=50 \times 50=2500 \mathrm{~V}$ And $\quad \mathrm{I}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{50}{10}=5 \mathrm{~A}$ $\therefore \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{5 \times 50}{2500}=\frac{1}{10}$ $\mathrm{I}_{1}=\frac{1}{10} \mathrm{~A}$
155389
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the ratio of output to input power is 0.8 then, the current drawn by primary winding is
1 $2.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $3.6 \mathrm{~A}$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}$ $\mathrm{~V}_{\mathrm{S}}=440 \mathrm{~V}$ $\mathrm{I}_{\mathrm{S}}=2.0 \mathrm{~A}$ $\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{I}}}=0.8$ We know that, Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{P}_{\mathrm{O}}}{\mathrm{P}_{\mathrm{I}}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}$ $0.8=\frac{440 \times 2.0}{220 \times \mathrm{I}_{\mathrm{P}}}$ $\mathrm{I}_{\mathrm{P}}=\frac{2 \times 2.0}{0.8}$ $\mathrm{I}_{\mathrm{P}}=5.0 \mathrm{~A}$
MHT-CET 2019
Alternating Current
155390
The input voltage of an ideal transformer is $1000 \mathrm{~V}$ and input current is $50 \mathrm{~A}$. If the output voltage is $220 \mathrm{~V}$, then find the resistance of secondary coil.
1 $2 \Omega$
2 $3 \Omega$
3 $1 \Omega$
4 $4 \Omega$
Explanation:
C Given, $\mathrm{V}_{\mathrm{P}}=1000 \mathrm{~V}, \mathrm{I}_{\mathrm{P}}=50 \mathrm{~A}, \mathrm{~V}_{\mathrm{S}}=220 \mathrm{~V}$ We know that, for an ideal transformer- $\mathrm{P}_{\text {in }} =\mathrm{P}_{\text {out }}$ $\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} =\frac{\mathrm{V}_{\mathrm{s}}^{2}}{\mathrm{R}_{\mathrm{s}}}$ Putting these value, we get- $1000 \times 50=\frac{(220)^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{484}{500}=.968 \approx 1 \Omega$
AIIMS-26.05.2019(E) Shift-2
Alternating Current
155391
A transformer with turns ratio $\frac{N_{1}}{N_{2}}=\frac{50}{1}$ is connected to a 120 volt $\mathrm{AC}$ supply. If primary and secondary circuit resistance are $1.5 \mathrm{k} \Omega$ and $1 \Omega$ respectively then find out power output.
1 $5.76 \mathrm{~W}$
2 $11.4 \mathrm{~W}$
3 $2.89 \mathrm{~W}$
4 $7.56 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=1.5 \mathrm{~K} \Omega$ $\mathrm{R}_{2}=1 \Omega$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{50}{1}$ $\mathrm{V}_{1}=120$ Volts We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Putting these value, we get- $\frac{50}{1}=\frac{120}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{12}{5} \mathrm{Volt}$ $\text { And } \quad \mathrm{P}_{\text {output }}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}}=\frac{\left(\frac{12}{5}\right)^{2}}{1}=\frac{144}{25}=5.76 \mathrm{~W}$
AIIMS-25.05.2019(M) Shift-1
Alternating Current
155392
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 \mathrm{~A}$, then that in the secondary is.
1 $4 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{\mathrm{P}}\right)=140$ Number of turns secondary coil $\left(\mathrm{N}_{\mathrm{S}}\right)=280$ Primary current $\left(\mathrm{I}_{\mathrm{P}}\right)=4 \mathrm{~A}$ We know that, Transformer ratio, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{N_{P}}{N_{S}}=\frac{I_{S}}{I_{P}}$ $\frac{140}{280}=\frac{I_{s}}{4}$ $I_{S}=2 \mathrm{~A}$
AIIMS-25.05.2019(E) Shift-2
Alternating Current
155393
A transformer consists of 500 turn in primary coil and $\mathbf{1 0}$ turns in secondary coil with the load of $10 \Omega$. Find out current in the primary coil when the voltage across secondary coil is $50 \mathrm{~V}$.
1 $5 \mathrm{~A}$
2 $\frac{1}{10} \mathrm{~A}$
3 $10 \mathrm{~A}$
4 $\frac{1}{20} \mathrm{~A}$
Explanation:
B Given, $\mathrm{N}_{1}=500$ turn $\mathrm{N}_{2}=10$ turn $\mathrm{V}_{2}=50$ volt We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Or $\quad \mathrm{V}_{1}=50 \times 50=2500 \mathrm{~V}$ And $\quad \mathrm{I}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{50}{10}=5 \mathrm{~A}$ $\therefore \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{5 \times 50}{2500}=\frac{1}{10}$ $\mathrm{I}_{1}=\frac{1}{10} \mathrm{~A}$
