NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155382
A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are
155385
The primary of a transformer has 100 turns and operates at $100 \mathrm{~V}-200 \mathrm{~W}$. The number of turns in the secondary, if the output voltage is $2000 \mathrm{~V}$, then
1 2000
2 200
3 100
4 500
Explanation:
A Given that, Primary turns $\left(\mathrm{N}_{\mathrm{P}}\right)=100$ Primary voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{S}}\right)=2000\mathrm{~V}$ From equation of transformer, $\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}$ $\mathrm{N}_{\mathrm{S}}=\frac{\mathrm{N}_{\mathrm{P}} \times \mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{100 \times 2000}{100}$ $\mathrm{~N}_{\mathrm{S}}=2000$
AP EAMCET (21.09.2020) Shift-II
Alternating Current
155386
Find the current through the primary coil $(P)$ of the transformer shown below.
1 $0.08 \mathrm{~A}$
2 $0.04 \mathrm{~A}$
3 $0.02 \mathrm{~A}$
4 $0.01 \mathrm{~A}$
Explanation:
C Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$, Secondary voltage $\left(\mathrm{V}_{2}\right)=23 \mathrm{~V}$, Resistance $(\mathrm{R})=115 \Omega$ Current in secondary $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{23}{115}$ From voltage current relation in transformer $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{23}{115} \times \frac{23}{230}$ $\mathrm{I}_{1}=0.02 \mathrm{~A}$
AP EAMCET (21.09.2020) Shift-I
Alternating Current
155387
An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 \mathrm{~V}, 8$ $A$, under these conditions. What would the meters of the primary read?
1 $100 \mathrm{~V}, 16 \mathrm{~A}$
2 $40 \mathrm{~V}, 40 \mathrm{~A}$
3 $160 \mathrm{~V}, 10 \mathrm{~A}$
4 $80 \mathrm{~V}, 20 \mathrm{~A}$
Explanation:
B Given that, Primary turns $\left(\mathrm{N}_{1}\right)=500$ Secondary turns $\left(\mathrm{N}_{2}\right)=2500$ Secondary voltage $\left(\mathrm{V}_{2}\right)=200 \mathrm{~V}$ Current in secondary $\left(\mathrm{I}_{2}\right)=8 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{500}{2500}=\frac{\mathrm{V}_{1}}{200}$ $\mathrm{~V}_{1}=\frac{500 \times 200}{2500}=40 \mathrm{~V}$ And $\quad \frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{40}{200}=\frac{8}{I_{1}}$ $I_{1}=\frac{8 \times 200}{40}=40 \mathrm{~A}$ $I_{1}=40 \mathrm{~A}$ Hence, the reading on primary meter is $40 \mathrm{~V}, 40 \mathrm{~A}$.
155382
A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are
155385
The primary of a transformer has 100 turns and operates at $100 \mathrm{~V}-200 \mathrm{~W}$. The number of turns in the secondary, if the output voltage is $2000 \mathrm{~V}$, then
1 2000
2 200
3 100
4 500
Explanation:
A Given that, Primary turns $\left(\mathrm{N}_{\mathrm{P}}\right)=100$ Primary voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{S}}\right)=2000\mathrm{~V}$ From equation of transformer, $\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}$ $\mathrm{N}_{\mathrm{S}}=\frac{\mathrm{N}_{\mathrm{P}} \times \mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{100 \times 2000}{100}$ $\mathrm{~N}_{\mathrm{S}}=2000$
AP EAMCET (21.09.2020) Shift-II
Alternating Current
155386
Find the current through the primary coil $(P)$ of the transformer shown below.
1 $0.08 \mathrm{~A}$
2 $0.04 \mathrm{~A}$
3 $0.02 \mathrm{~A}$
4 $0.01 \mathrm{~A}$
Explanation:
C Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$, Secondary voltage $\left(\mathrm{V}_{2}\right)=23 \mathrm{~V}$, Resistance $(\mathrm{R})=115 \Omega$ Current in secondary $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{23}{115}$ From voltage current relation in transformer $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{23}{115} \times \frac{23}{230}$ $\mathrm{I}_{1}=0.02 \mathrm{~A}$
AP EAMCET (21.09.2020) Shift-I
Alternating Current
155387
An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 \mathrm{~V}, 8$ $A$, under these conditions. What would the meters of the primary read?
1 $100 \mathrm{~V}, 16 \mathrm{~A}$
2 $40 \mathrm{~V}, 40 \mathrm{~A}$
3 $160 \mathrm{~V}, 10 \mathrm{~A}$
4 $80 \mathrm{~V}, 20 \mathrm{~A}$
Explanation:
B Given that, Primary turns $\left(\mathrm{N}_{1}\right)=500$ Secondary turns $\left(\mathrm{N}_{2}\right)=2500$ Secondary voltage $\left(\mathrm{V}_{2}\right)=200 \mathrm{~V}$ Current in secondary $\left(\mathrm{I}_{2}\right)=8 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{500}{2500}=\frac{\mathrm{V}_{1}}{200}$ $\mathrm{~V}_{1}=\frac{500 \times 200}{2500}=40 \mathrm{~V}$ And $\quad \frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{40}{200}=\frac{8}{I_{1}}$ $I_{1}=\frac{8 \times 200}{40}=40 \mathrm{~A}$ $I_{1}=40 \mathrm{~A}$ Hence, the reading on primary meter is $40 \mathrm{~V}, 40 \mathrm{~A}$.
155382
A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are
155385
The primary of a transformer has 100 turns and operates at $100 \mathrm{~V}-200 \mathrm{~W}$. The number of turns in the secondary, if the output voltage is $2000 \mathrm{~V}$, then
1 2000
2 200
3 100
4 500
Explanation:
A Given that, Primary turns $\left(\mathrm{N}_{\mathrm{P}}\right)=100$ Primary voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{S}}\right)=2000\mathrm{~V}$ From equation of transformer, $\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}$ $\mathrm{N}_{\mathrm{S}}=\frac{\mathrm{N}_{\mathrm{P}} \times \mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{100 \times 2000}{100}$ $\mathrm{~N}_{\mathrm{S}}=2000$
AP EAMCET (21.09.2020) Shift-II
Alternating Current
155386
Find the current through the primary coil $(P)$ of the transformer shown below.
1 $0.08 \mathrm{~A}$
2 $0.04 \mathrm{~A}$
3 $0.02 \mathrm{~A}$
4 $0.01 \mathrm{~A}$
Explanation:
C Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$, Secondary voltage $\left(\mathrm{V}_{2}\right)=23 \mathrm{~V}$, Resistance $(\mathrm{R})=115 \Omega$ Current in secondary $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{23}{115}$ From voltage current relation in transformer $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{23}{115} \times \frac{23}{230}$ $\mathrm{I}_{1}=0.02 \mathrm{~A}$
AP EAMCET (21.09.2020) Shift-I
Alternating Current
155387
An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 \mathrm{~V}, 8$ $A$, under these conditions. What would the meters of the primary read?
1 $100 \mathrm{~V}, 16 \mathrm{~A}$
2 $40 \mathrm{~V}, 40 \mathrm{~A}$
3 $160 \mathrm{~V}, 10 \mathrm{~A}$
4 $80 \mathrm{~V}, 20 \mathrm{~A}$
Explanation:
B Given that, Primary turns $\left(\mathrm{N}_{1}\right)=500$ Secondary turns $\left(\mathrm{N}_{2}\right)=2500$ Secondary voltage $\left(\mathrm{V}_{2}\right)=200 \mathrm{~V}$ Current in secondary $\left(\mathrm{I}_{2}\right)=8 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{500}{2500}=\frac{\mathrm{V}_{1}}{200}$ $\mathrm{~V}_{1}=\frac{500 \times 200}{2500}=40 \mathrm{~V}$ And $\quad \frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{40}{200}=\frac{8}{I_{1}}$ $I_{1}=\frac{8 \times 200}{40}=40 \mathrm{~A}$ $I_{1}=40 \mathrm{~A}$ Hence, the reading on primary meter is $40 \mathrm{~V}, 40 \mathrm{~A}$.
155382
A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are
155385
The primary of a transformer has 100 turns and operates at $100 \mathrm{~V}-200 \mathrm{~W}$. The number of turns in the secondary, if the output voltage is $2000 \mathrm{~V}$, then
1 2000
2 200
3 100
4 500
Explanation:
A Given that, Primary turns $\left(\mathrm{N}_{\mathrm{P}}\right)=100$ Primary voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{S}}\right)=2000\mathrm{~V}$ From equation of transformer, $\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}$ $\mathrm{N}_{\mathrm{S}}=\frac{\mathrm{N}_{\mathrm{P}} \times \mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{100 \times 2000}{100}$ $\mathrm{~N}_{\mathrm{S}}=2000$
AP EAMCET (21.09.2020) Shift-II
Alternating Current
155386
Find the current through the primary coil $(P)$ of the transformer shown below.
1 $0.08 \mathrm{~A}$
2 $0.04 \mathrm{~A}$
3 $0.02 \mathrm{~A}$
4 $0.01 \mathrm{~A}$
Explanation:
C Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$, Secondary voltage $\left(\mathrm{V}_{2}\right)=23 \mathrm{~V}$, Resistance $(\mathrm{R})=115 \Omega$ Current in secondary $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{23}{115}$ From voltage current relation in transformer $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{\mathrm{I}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}}=\frac{23}{115} \times \frac{23}{230}$ $\mathrm{I}_{1}=0.02 \mathrm{~A}$
AP EAMCET (21.09.2020) Shift-I
Alternating Current
155387
An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 \mathrm{~V}, 8$ $A$, under these conditions. What would the meters of the primary read?
1 $100 \mathrm{~V}, 16 \mathrm{~A}$
2 $40 \mathrm{~V}, 40 \mathrm{~A}$
3 $160 \mathrm{~V}, 10 \mathrm{~A}$
4 $80 \mathrm{~V}, 20 \mathrm{~A}$
Explanation:
B Given that, Primary turns $\left(\mathrm{N}_{1}\right)=500$ Secondary turns $\left(\mathrm{N}_{2}\right)=2500$ Secondary voltage $\left(\mathrm{V}_{2}\right)=200 \mathrm{~V}$ Current in secondary $\left(\mathrm{I}_{2}\right)=8 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{500}{2500}=\frac{\mathrm{V}_{1}}{200}$ $\mathrm{~V}_{1}=\frac{500 \times 200}{2500}=40 \mathrm{~V}$ And $\quad \frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{40}{200}=\frac{8}{I_{1}}$ $I_{1}=\frac{8 \times 200}{40}=40 \mathrm{~A}$ $I_{1}=40 \mathrm{~A}$ Hence, the reading on primary meter is $40 \mathrm{~V}, 40 \mathrm{~A}$.
