356334
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are:
We have, \(\begin{aligned}& \dfrac{N_{s}}{N_{p}}=\dfrac{I_{p}}{I_{s}}, \dfrac{10}{500}=\dfrac{I_{p}}{I_{s}} \Rightarrow \dfrac{I_{p}}{I_{s}}=\dfrac{1}{50} \\& \Rightarrow I_{s}=50 I_{p}\end{aligned}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356335
\(A{\rm{ }}12\;V,60\;W\) lamp is connected to the secondary of a step down transformer, whose primary is connected to \(ac\) mains of \(220\;V\). Assuming the transformer to be ideal, what is the current in the primary winding?
356336
A step up transformer has 300 turns of primary winding and 450 turns of secondary winding . A primary is connected to 150 \(V\) and the current flowing through it is 9 \(A\). The current and voltage in the secondary are
1 \(13.5\,A,100\,V\)
2 \(13.5\,A,225\,V\)
3 \(4.5\,A,100\,V\)
4 \(6.0\,A,225\,V\)
Explanation:
Given, number of turns in primary winding, \({N_P} = 300\). Number of turns in secondary winding, \({N_S} = 450\). Primary voltage,\({V_P} = 150V\). Primary current, \({I_P} = 9A\). For step - up transformer \(\frac{{{V_S}}}{{{V_P}}} = \frac{{{N_S}}}{{{N_P}}}\) \( \Rightarrow \frac{{{V_S}}}{{150}} = \frac{{450}}{{300}}\) \( \Rightarrow {V_S} = \frac{{450}}{{300}} \times 150 = 225V\) Again,\({V_P}{I_P} = {V_S}{I_S}\) (assuming ideal transformer) \( \Rightarrow 150 \times 9 = 225 \times {I_S}\) \( \Rightarrow {I_S} = \frac{{1350}}{{225}} = 6.0A\)
MHTCET - 2020
PHXII07:ALTERNATING CURRENT
356337
Turn ratio is \(1: 25\). The step up transformer operates at \(230\;V\)and current through secondary is \(2\;A\). Then current in primary is
1 \(25\;A\)
2 \(100\;A\)
3 \(50\;A\)
4 \(20\;A\)
Explanation:
Here, \(\frac{{{N_P}}}{{{N_S}}} = \frac{1}{{25}},{V_P} = 230\;V,{I_S} = 2\;A\) For an ideal transformer \(\dfrac{N_{P}}{N_{S}}=\dfrac{V_{P}}{I_{S}}=\dfrac{I_{S}}{I_{P}} \quad\) or \(\dfrac{N_{P}}{N_{S}}=\dfrac{I_{S}}{I_{P}}\) or\({I_P} = {I_S} \times \frac{{{N_S}}}{{{N_P}}} = 2\;A \times \frac{{25}}{1} = 50\;A\)
356334
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are:
We have, \(\begin{aligned}& \dfrac{N_{s}}{N_{p}}=\dfrac{I_{p}}{I_{s}}, \dfrac{10}{500}=\dfrac{I_{p}}{I_{s}} \Rightarrow \dfrac{I_{p}}{I_{s}}=\dfrac{1}{50} \\& \Rightarrow I_{s}=50 I_{p}\end{aligned}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356335
\(A{\rm{ }}12\;V,60\;W\) lamp is connected to the secondary of a step down transformer, whose primary is connected to \(ac\) mains of \(220\;V\). Assuming the transformer to be ideal, what is the current in the primary winding?
356336
A step up transformer has 300 turns of primary winding and 450 turns of secondary winding . A primary is connected to 150 \(V\) and the current flowing through it is 9 \(A\). The current and voltage in the secondary are
1 \(13.5\,A,100\,V\)
2 \(13.5\,A,225\,V\)
3 \(4.5\,A,100\,V\)
4 \(6.0\,A,225\,V\)
Explanation:
Given, number of turns in primary winding, \({N_P} = 300\). Number of turns in secondary winding, \({N_S} = 450\). Primary voltage,\({V_P} = 150V\). Primary current, \({I_P} = 9A\). For step - up transformer \(\frac{{{V_S}}}{{{V_P}}} = \frac{{{N_S}}}{{{N_P}}}\) \( \Rightarrow \frac{{{V_S}}}{{150}} = \frac{{450}}{{300}}\) \( \Rightarrow {V_S} = \frac{{450}}{{300}} \times 150 = 225V\) Again,\({V_P}{I_P} = {V_S}{I_S}\) (assuming ideal transformer) \( \Rightarrow 150 \times 9 = 225 \times {I_S}\) \( \Rightarrow {I_S} = \frac{{1350}}{{225}} = 6.0A\)
MHTCET - 2020
PHXII07:ALTERNATING CURRENT
356337
Turn ratio is \(1: 25\). The step up transformer operates at \(230\;V\)and current through secondary is \(2\;A\). Then current in primary is
1 \(25\;A\)
2 \(100\;A\)
3 \(50\;A\)
4 \(20\;A\)
Explanation:
Here, \(\frac{{{N_P}}}{{{N_S}}} = \frac{1}{{25}},{V_P} = 230\;V,{I_S} = 2\;A\) For an ideal transformer \(\dfrac{N_{P}}{N_{S}}=\dfrac{V_{P}}{I_{S}}=\dfrac{I_{S}}{I_{P}} \quad\) or \(\dfrac{N_{P}}{N_{S}}=\dfrac{I_{S}}{I_{P}}\) or\({I_P} = {I_S} \times \frac{{{N_S}}}{{{N_P}}} = 2\;A \times \frac{{25}}{1} = 50\;A\)
356334
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are:
We have, \(\begin{aligned}& \dfrac{N_{s}}{N_{p}}=\dfrac{I_{p}}{I_{s}}, \dfrac{10}{500}=\dfrac{I_{p}}{I_{s}} \Rightarrow \dfrac{I_{p}}{I_{s}}=\dfrac{1}{50} \\& \Rightarrow I_{s}=50 I_{p}\end{aligned}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356335
\(A{\rm{ }}12\;V,60\;W\) lamp is connected to the secondary of a step down transformer, whose primary is connected to \(ac\) mains of \(220\;V\). Assuming the transformer to be ideal, what is the current in the primary winding?
356336
A step up transformer has 300 turns of primary winding and 450 turns of secondary winding . A primary is connected to 150 \(V\) and the current flowing through it is 9 \(A\). The current and voltage in the secondary are
1 \(13.5\,A,100\,V\)
2 \(13.5\,A,225\,V\)
3 \(4.5\,A,100\,V\)
4 \(6.0\,A,225\,V\)
Explanation:
Given, number of turns in primary winding, \({N_P} = 300\). Number of turns in secondary winding, \({N_S} = 450\). Primary voltage,\({V_P} = 150V\). Primary current, \({I_P} = 9A\). For step - up transformer \(\frac{{{V_S}}}{{{V_P}}} = \frac{{{N_S}}}{{{N_P}}}\) \( \Rightarrow \frac{{{V_S}}}{{150}} = \frac{{450}}{{300}}\) \( \Rightarrow {V_S} = \frac{{450}}{{300}} \times 150 = 225V\) Again,\({V_P}{I_P} = {V_S}{I_S}\) (assuming ideal transformer) \( \Rightarrow 150 \times 9 = 225 \times {I_S}\) \( \Rightarrow {I_S} = \frac{{1350}}{{225}} = 6.0A\)
MHTCET - 2020
PHXII07:ALTERNATING CURRENT
356337
Turn ratio is \(1: 25\). The step up transformer operates at \(230\;V\)and current through secondary is \(2\;A\). Then current in primary is
1 \(25\;A\)
2 \(100\;A\)
3 \(50\;A\)
4 \(20\;A\)
Explanation:
Here, \(\frac{{{N_P}}}{{{N_S}}} = \frac{1}{{25}},{V_P} = 230\;V,{I_S} = 2\;A\) For an ideal transformer \(\dfrac{N_{P}}{N_{S}}=\dfrac{V_{P}}{I_{S}}=\dfrac{I_{S}}{I_{P}} \quad\) or \(\dfrac{N_{P}}{N_{S}}=\dfrac{I_{S}}{I_{P}}\) or\({I_P} = {I_S} \times \frac{{{N_S}}}{{{N_P}}} = 2\;A \times \frac{{25}}{1} = 50\;A\)
356334
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are:
We have, \(\begin{aligned}& \dfrac{N_{s}}{N_{p}}=\dfrac{I_{p}}{I_{s}}, \dfrac{10}{500}=\dfrac{I_{p}}{I_{s}} \Rightarrow \dfrac{I_{p}}{I_{s}}=\dfrac{1}{50} \\& \Rightarrow I_{s}=50 I_{p}\end{aligned}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356335
\(A{\rm{ }}12\;V,60\;W\) lamp is connected to the secondary of a step down transformer, whose primary is connected to \(ac\) mains of \(220\;V\). Assuming the transformer to be ideal, what is the current in the primary winding?
356336
A step up transformer has 300 turns of primary winding and 450 turns of secondary winding . A primary is connected to 150 \(V\) and the current flowing through it is 9 \(A\). The current and voltage in the secondary are
1 \(13.5\,A,100\,V\)
2 \(13.5\,A,225\,V\)
3 \(4.5\,A,100\,V\)
4 \(6.0\,A,225\,V\)
Explanation:
Given, number of turns in primary winding, \({N_P} = 300\). Number of turns in secondary winding, \({N_S} = 450\). Primary voltage,\({V_P} = 150V\). Primary current, \({I_P} = 9A\). For step - up transformer \(\frac{{{V_S}}}{{{V_P}}} = \frac{{{N_S}}}{{{N_P}}}\) \( \Rightarrow \frac{{{V_S}}}{{150}} = \frac{{450}}{{300}}\) \( \Rightarrow {V_S} = \frac{{450}}{{300}} \times 150 = 225V\) Again,\({V_P}{I_P} = {V_S}{I_S}\) (assuming ideal transformer) \( \Rightarrow 150 \times 9 = 225 \times {I_S}\) \( \Rightarrow {I_S} = \frac{{1350}}{{225}} = 6.0A\)
MHTCET - 2020
PHXII07:ALTERNATING CURRENT
356337
Turn ratio is \(1: 25\). The step up transformer operates at \(230\;V\)and current through secondary is \(2\;A\). Then current in primary is
1 \(25\;A\)
2 \(100\;A\)
3 \(50\;A\)
4 \(20\;A\)
Explanation:
Here, \(\frac{{{N_P}}}{{{N_S}}} = \frac{1}{{25}},{V_P} = 230\;V,{I_S} = 2\;A\) For an ideal transformer \(\dfrac{N_{P}}{N_{S}}=\dfrac{V_{P}}{I_{S}}=\dfrac{I_{S}}{I_{P}} \quad\) or \(\dfrac{N_{P}}{N_{S}}=\dfrac{I_{S}}{I_{P}}\) or\({I_P} = {I_S} \times \frac{{{N_S}}}{{{N_P}}} = 2\;A \times \frac{{25}}{1} = 50\;A\)
