Nice — let’s solve this carefully, step-by-step.
**Given:** an ideal single-phase transformer with turns ratio (1:5) (primary : secondary).
Primary is connected to a (25\ \text{kV}) source. The secondary feeds a (30\ \Omega) resistor. Use a (100\ \text{MVA}) base for the per-unit part.
---
## (a) Direct (no per-unit) — load power and source current
1. **Voltage on secondary** (turns ratio (a = N_s/N_p = 5)):
[
V_s = a,V_p = 5\times 25,\text{kV} = 125,\text{kV}.
]
2. **Secondary current** (resistive load):
[
I_s = \frac{V_s}{R} = \frac{125{,}000}{30} = 4166.6667\ \text{A} = 4.16667\ \text{kA}.
]
3. **Power delivered to load** (apparent = real since purely resistive):
[
S_{\text{load}} = V_s I_s = \frac{V_s^2}{R} = 125{,}000\times 4166.6667
=520{,}833{,}333\ \text{VA}=520.833333\ \text{MVA}.
]
4. **Primary (source) current** — for an ideal transformer currents scale by the turns ratio:
[
I_p = \frac{N_s}{N_p},I_s = a,I_s = 5\times 4166.6667 =20{,}833.3333\ \text{A}=20.833333\ \text{kA}.
]
**Answers (a):**
* Load power = **520.833 MVA**
* Source current = **20.833 kA**
(You may round as needed, e.g. 520.83 MVA and 20.83 kA.)
---
## (b) Per-unit base table (use (S_b=100\ \text{MVA}))
We take voltage bases equal to the actual nominal voltages on each side.
* (S_b = 100\ \text{MVA}) (both sides)
* (V_{b,\text{primary}} = 25\ \text{kV})
* (V_{b,\text{secondary}} = 125\ \text{kV})
Compute base currents and base impedances (single-phase: (I_b = S_b / V_b), (Z_b = V_b^2/S_b)).
Primary side:
* (S_b = 100\ \text{MVA})
* (V_b = 25\ \text{kV})
* (I_b = \dfrac{100\times10^6}{25\times10^3} = 4000\ \text{A})
* (Z_b = \dfrac{(25\times10^3)^2}{100\times10^6} = 6.25\ \Omega)
Secondary side:
* (S_b = 100\ \text{MVA})
* (V_b = 125\ \text{kV})
* (I_b = \dfrac{100\times10^6}{125\times10^3} = 800\ \text{A})
* (Z_b = \dfrac{(125\times10^3)^2}{100\times10^6} = 156.25\ \Omega)
**Table (b):**
| Side | (S_b) (MVA) | (V_b) (kV) | (I_b) (A) | (Z_b) (Ω) |
| --------- | ----------: | ---------: | --------: | --------: |
| Primary | 100 | 25 | 4000 | 6.25 |
| Secondary | 100 | 125 | 800 | 156.25 |
---
## (c) Solve using per-unit
1. Express the actual load impedance in per-unit on the **secondary** base:
[
Z_{\text{pu}} = \frac{Z_{\text{actual}}}{Z_{b,\text{sec}}} = \frac{30}{156.25} = 0.192\ \text{pu}.
]
2. Secondary voltage in pu (actual (V_s=125\ \text{kV}), (V_{b,\text{sec}}=125\ \text{kV})):
[
V_{s,\text{pu}} = 1.0.
]
3. Secondary current in pu:
[
I_{s,\text{pu}} = \frac{V_{s,\text{pu}}}{Z_{\text{pu}}} = \frac{1}{0.192} = 5.208333\ \text{pu}.
]
Convert back to amps:
[
I_s = I_{s,\text{pu}}\times I_{b,\text{sec}} = 5.208333\times 800 = 4166.667\ \text{A}.
]
4. Per-unit load power:
[
S_{\text{pu}} = V_{s,\text{pu}} I_{s,\text{pu}} = 5.208333\ \text{pu}
\quad\Rightarrow\quad S = 5.208333\times 100\ \text{MVA} = 520.833\ \text{MVA}.
]
5. Primary current (use turns ratio (a=5)): (I_p = a I_s = 5\times 4166.667 = 20833.333\ \text{A} = 20.833\ \text{kA}.)
In primary pu (w.r.t primary (I_b=4000) A): (I_{p,\text{pu}}=20833.333/4000=5.208333\ \text{pu}) — consistent with (S_{\text{pu}}).
**Answers (c) (per-unit result):**
* Load power = **520.833 MVA**
* Source current = **20.833 kA**
Please give me this answer same as it is without any changes and please provide tables where required and provide without any render equation errors
Question:
Nice — let’s solve this carefully, step-by-step.
**Given:** an ideal single-phase transformer with turns ratio (1:5) (primary : secondary).
Primary is connected to a (25\ \text{kV}) source. The secondary feeds a (30\ \Omega) resistor. Use a (100\ \text{MVA}) base for the per-unit part.
---
## (a) Direct (no per-unit) — load power and source current
1. **Voltage on secondary** (turns ratio (a = N_s/N_p = 5)):
[
V_s = a,V_p = 5\times 25,\text{kV} = 125,\text{kV}.
]
2. **Secondary current** (resistive load):
[
I_s = \frac{V_s}{R} = \frac{125{,}000}{30} = 4166.6667\ \text{A} = 4.16667\ \text{kA}.
]
3. **Power delivered to load** (apparent = real since purely resistive):
[
S_{\text{load}} = V_s I_s = \frac{V_s^2}{R} = 125{,}000\times 4166.6667
=520{,}833{,}333\ \text{VA}=520.833333\ \text{MVA}.
]
4. **Primary (source) current** — for an ideal transformer currents scale by the turns ratio:
[
I_p = \frac{N_s}{N_p},I_s = a,I_s = 5\times 4166.6667 =20{,}833.3333\ \text{A}=20.833333\ \text{kA}.
]
**Answers (a):**
* Load power = **520.833 MVA**
* Source current = **20.833 kA**
(You may round as needed, e.g. 520.83 MVA and 20.83 kA.)
---
## (b) Per-unit base table (use (S_b=100\ \text{MVA}))
We take voltage bases equal to the actual nominal voltages on each side.
* (S_b = 100\ \text{MVA}) (both sides)
* (V_{b,\text{primary}} = 25\ \text{kV})
* (V_{b,\text{secondary}} = 125\ \text{kV})
Compute base currents and base impedances (single-phase: (I_b = S_b / V_b), (Z_b = V_b^2/S_b)).
Primary side:
* (S_b = 100\ \text{MVA})
* (V_b = 25\ \text{kV})
* (I_b = \dfrac{100\times10^6}{25\times10^3} = 4000\ \text{A})
* (Z_b = \dfrac{(25\times10^3)^2}{100\times10^6} = 6.25\ \Omega)
Secondary side:
* (S_b = 100\ \text{MVA})
* (V_b = 125\ \text{kV})
* (I_b = \dfrac{100\times10^6}{125\times10^3} = 800\ \text{A})
* (Z_b = \dfrac{(125\times10^3)^2}{100\times10^6} = 156.25\ \Omega)
**Table (b):**
| Side | (S_b) (MVA) | (V_b) (kV) | (I_b) (A) | (Z_b) (Ω) |
| --------- | ----------: | ---------: | --------: | --------: |
| Primary | 100 | 25 | 4000 | 6.25 |
| Secondary | 100 | 125 | 800 | 156.25 |
---
## (c) Solve using per-unit
1. Express the actual load impedance in per-unit on the **secondary** base:
[
Z_{\text{pu}} = \frac{Z_{\text{actual}}}{Z_{b,\text{sec}}} = \frac{30}{156.25} = 0.192\ \text{pu}.
]
2. Secondary voltage in pu (actual (V_s=125\ \text{kV}), (V_{b,\text{sec}}=125\ \text{kV})):
[
V_{s,\text{pu}} = 1.0.
]
3. Secondary current in pu:
[
I_{s,\text{pu}} = \frac{V_{s,\text{pu}}}{Z_{\text{pu}}} = \frac{1}{0.192} = 5.208333\ \text{pu}.
]
Convert back to amps:
[
I_s = I_{s,\text{pu}}\times I_{b,\text{sec}} = 5.208333\times 800 = 4166.667\ \text{A}.
]
4. Per-unit load power:
[
S_{\text{pu}} = V_{s,\text{pu}} I_{s,\text{pu}} = 5.208333\ \text{pu}
\quad\Rightarrow\quad S = 5.208333\times 100\ \text{MVA} = 520.833\ \text{MVA}.
]
5. Primary current (use turns ratio (a=5)): (I_p = a I_s = 5\times 4166.667 = 20833.333\ \text{A} = 20.833\ \text{kA}.)
In primary pu (w.r.t primary (I_b=4000) A): (I_{p,\text{pu}}=20833.333/4000=5.208333\ \text{pu}) — consistent with (S_{\text{pu}}).
**Answers (c) (per-unit result):**
* Load power = **520.833 MVA**
* Source current = **20.833 kA**
Please give me this answer same as it is without any changes and please provide tables where required and provide without any render equation errors
This is AI generated Response. Asked by: Mohin Created at: 2025-10-19 07:07:50
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Nice — let’s solve this carefully, step-by-step.
# (a) Direct (no per-unit) load power and source
1. **Voltage secondary** (turns \(a = N_s/N_p 5\)):
\[
V_s = atimes V_p 5 \ 25\,\text{kV = 125\,\text{kV}
\]
2. **Secondary** (resistive load):
[
I_s \frac{_s}{R} = \frac{125,000\text{}}{30\,\Omega} \approx 4166.6667\,\text{A} \approx 4.16667\,\text{kA}
\]
3. **Power delivered to load** (since purely resistive):
\[
S_{\text{load}} = V_s I_s = \frac{V_s^2}{R} = 125,000^2 / 30 \approx 520,833,333\,\text{VA} = 520.833333\,\text{MVA}
\]
4. **Primary (source) current** — for an ideal transformer:
\[
I_p = \frac{N_s}{N_p} \times I_s = 5 \times 4166.6667\,\text{A} = 20,833.3333\,\text{A} \approx 20.83\,\text{kA}
\]
---
### **Answers (a):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
---
# (b) Per-unit base table (use \(S_b=100\,\text{MVA}\))
Choose voltage bases:
- Primary: \(V_{b,\text{primary}} = 25\,\text{kV}\)
- Secondary: \(V_{b,\text{secondary}} = 125\,\text{kV}\)
Compute base currents and base impedances:
\[
I_b = \frac{S_b}{V_b}
\]
\[
Z_b = \frac{V_b^2}{S_b}
\]
### Primary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{primary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{25 \times 10^3\,\text{V}}\) | 400 A |
| \(Z_{b,\text{primary}}\) | \(\frac{(25 \times 10^3)^2}{100 \times 10^6}\) | 6.25 Ω |
### Secondary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{secondary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{125 \times 10^3\,\text{V}}\) | 800 A |
| \(Z_{b,\text{secondary}}\) | \(\frac{(125 \times 10^3)^2}{100 \times 10^6}\) | 156.25 Ω |
### **Table (b):**
| Side | \(S_b\) (MVA) | \(V_b\) (kV) | \(I_b\) (A) | \(Z_b\) (Ω) |
|-----------|--------------|--------------|--------------|--------------|
| Primary | 100 | 25 | 400 | 6.25 |
| Secondary | 100 | 125 | 800 | 156.25 |
---
# (c) Solve using per-unit
1. **Convert load impedance to per-unit on secondary base:**
\[
Z_{\text{actual}} = 30\,\Omega
\]
\[
Z_{b,\text{sec}} = 156.25\,\Omega
\]
\[
Z_{\text{pu}} = \frac{Z_{\text{actual}}}{Z_{b,\text{sec}}} = \frac{30}{156.25} \approx .192
\]
2. **Secondary voltage in per-unit:**
\[
V_{s,\text{pu}} = \frac{V_s}{V_{b,\text{sec}}} = \frac{125\,\text{kV}}{125\,\text{kV}} = 1.
\]
3. **Secondary current in per-unit:**
\[
I_{s,\text{pu}} = \frac{V_{s,\text{pu}}}{Z_{\text{pu}}} = \frac{1.}{.192} \approx 5.208333\,\text{pu}
\]
4. **Convert back to actual amps:**
\[
I_s = I_{s,\text{pu}} \times I_{b,\text{secondary}} = 5.208333 \times 800\,\text{A} \approx 4166.667\,\text{A}
\]
5. **Per-unit load power:**
\[
S_{\text{pu}} = V_{s,\text{pu}} \times I_{s,\text{pu}} = 1. \times 5.208333 = 5.208333\,\text{pu}
\]
Converted to actual:
\[
S = 5.208333 \times 100\,\text{MVA} = 520.833\,\text{MVA}
\]
6. **Primary current (using turns ratio \(a=5\))**:
\[
I_p = a \times I_s = 5 \times 4166.667\,\text{A} \approx 20,833.333\,\text{A}
\]
Expressed in primary per-unit (with \(I_{b,\text{primary}}=400\,\text{A}\)):
\[
I_{p,\text{pu}} = \frac{20,833.333}{400} \approx 5.208333\,\text{pu}
\]
---
### **Final answers (c):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
*Note:* The per-unit calculations confirm the direct calculation results.
---
Let me know if you'd like this formatted in a specific way or with additional details!
Question:
please give me the entire answer in render quations
This is AI generated Response. Asked by: Mohin Created at: 2025-10-19 07:10:07
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# (a) Direct (no per-unit) load power and source
1. **Voltage on secondary** (turns ratio \(a = \frac{N_s}{N_p} = 5\)):
\[
V_s = a \times V_p = 5 \times 25\,\text{kV} = 125\,\text{kV}
\]
2. **Secondary current** (resistive load):
\[
I_s = \frac{V_s}{R} = \frac{125{,}000\,\text{V}}{30\,\Omega} \approx 4166.6667\,\text{A} \approx 4.16667\,\text{kA}
\]
3. **Power delivered to load** (apparent = real since purely resistive):
\[
S_{\text{load}} = V_s \times I_s = \frac{V_s^2}{R} = 125{,}000 \times 4166.6667 \approx 520{,}833{,}333\,\text{VA} = 520.833333\,\text{MVA}
\]
4. **Primary (source) current** — for an ideal transformer:
\[
I_p = \frac{N_s}{N_p} \times I_s = 5 \times 4166.6667\,\text{A} \approx 20{,}833.3333\,\text{A} \approx 20.833\,\text{kA}
\]
---
### **Answers (a):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
---
# (b) Per-unit base table (use \(S_b=100\,\text{MVA}\))
Choose voltage bases:
- Primary: \(V_{b,\text{primary}} = 25\,\text{kV}\)
- Secondary: \(V_{b,\text{secondary}} = 125\,\text{kV}\)
Compute base currents and base impedances:
\[
I_b = \frac{S_b}{V_b}
\]
\[
Z_b = \frac{V_b^2}{S_b}
\]
### Primary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{primary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{25 \times 10^3\,\text{V}}\) | 4000 A |
| \(Z_{b,\text{primary}}\) | \(\frac{(25 \times 10^3)^2}{100 \times 10^6}\) | 6.25 Ω |
### Secondary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{secondary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{125 \times 10^3\,\text{V}}\) | 800 A |
| \(Z_{b,\text{secondary}}\) | \(\frac{(125 \times 10^3)^2}{100 \times 10^6}\) | 156.25 Ω |
### **Table (b):**
| Side | \(S_b\) (MVA) | \(V_b\) (kV) | \(I_b\) (A) | \(Z_b\) (Ω) |
|-----------|--------------|--------------|--------------|--------------|
| Primary | 100 | 25 | 4000 | 6.25 |
| Secondary | 100 | 125 | 800 | 156.25 |
---
# (c) Solve using per-unit
1. **Express the actual load impedance in per-unit on the secondary base:**
\[
Z_{\text{pu}} = \frac{Z_{\text{actual}}}{Z_{b,\text{sec}}} = \frac{30\,\Omega}{156.25\,\Omega} \approx 0.192\,\text{pu}
\]
2. **Secondary voltage in per-unit:**
\[
V_{s,\text{pu}} = \frac{V_s}{V_{b,\text{sec}}} = \frac{125\,\text{kV}}{125\,\text{kV}} = 1.0
\]
3. **Secondary current in per-unit:**
\[
I_{s,\text{pu}} = \frac{V_{s,\text{pu}}}{Z_{\text{pu}}} = \frac{1.0}{0.192} \approx 5.208333\,\text{pu}
\]
Convert back to amps:
\[
I_s = I_{s,\text{pu}} \times I_{b,\text{sec}} = 5.208333 \times 800\,\text{A} \approx 4166.667\,\text{A}
\]
4. **Per-unit load power:**
\[
S_{\text{pu}} = V_{s,\text{pu}} \times I_{s,\text{pu}} = 1.0 \times 5.208333 \approx 5.208333\,\text{pu}
\]
Convert back to actual:
\[
S = 5.208333 \times 100\,\text{MVA} \approx 520.833\,\text{MVA}
\]
5. **Primary current (use turns ratio \(a=5\))**:
\[
I_p = a \times I_s = 5 \times 4166.667\,\text{A} \approx 20{,}833.333\,\text{A} \approx 20.833\,\text{kA}
\]
In primary per-unit (with \(I_{b,\text{primary}}=4000\,\text{A}\)):
\[
I_{p,\text{pu}} = \frac{20{,}833.333}{4000} \approx 5.208333\,\text{pu}
\]
---
### **Final answers (c):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
*Note:* The per-unit calculations confirm the direct calculation results.
Question:
please give me the response clearly that it can be in copied format
This is AI generated Response. Asked by: Mohin Created at: 2025-10-19 07:21:35
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# (a) Direct (no per-unit) load power and source
1. **Voltage on secondary** (turns ratio \(a = \frac{N_s}{N_p} = 5\)):
\[
V_s = a \times V_p = 5 \times 25\,\text{kV} = 125\,\text{kV}
\]
2. **Secondary current** (resistive load):
\[
I_s = \frac{V_s}{R} = \frac{125{,}000\,\text{V}}{30\,\Omega} \approx 4166.6667\,\text{A} \approx 4.16667\,\text{kA}
\]
3. **Power delivered to load** (apparent = real since purely resistive):
\[
S_{\text{load}} = V_s \times I_s = \frac{V_s^2}{R} = 125{,}000 \times 4166.6667 \approx 520{,}833{,}333\,\text{VA} = 520.833333\,\text{MVA}
\]
4. **Primary (source) current** — for an ideal transformer:
\[
I_p = \frac{N_s}{N_p} \times I_s = 5 \times 4166.6667\,\text{A} \approx 20{,}833.3333\,\text{A} \approx 20.833\,\text{kA}
\]
---
### **Answers (a):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
---
# (b) Per-unit base table (use \(S_b=100\,\text{MVA}\))
Choose voltage bases:
- Primary: \(V_{b,\text{primary}} = 25\,\text{kV}\)
- Secondary: \(V_{b,\text{secondary}} = 125\,\text{kV}\)
Compute base currents and base impedances:
\[
I_b = \frac{S_b}{V_b}
\]
\[
Z_b = \frac{V_b^2}{S_b}
\]
### Primary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{primary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{25 \times 10^3\,\text{V}}\) | 4000 A |
| \(Z_{b,\text{primary}}\) | \(\frac{(25 \times 10^3)^2}{100 \times 10^6}\) | 6.25 Ω |
### Secondary side:
| Quantity | Calculation | Result |
|--------------|----------------------------------------------|----------------|
| \(I_{b,\text{secondary}}\) | \(\frac{100 \times 10^6\,\text{VA}}{125 \times 10^3\,\text{V}}\) | 800 A |
| \(Z_{b,\text{secondary}}\) | \(\frac{(125 \times 10^3)^2}{100 \times 10^6}\) | 156.25 Ω |
### **Table (b):**
| Side | \(S_b\) (MVA) | \(V_b\) (kV) | \(I_b\) (A) | \(Z_b\) (Ω) |
|-----------|--------------|--------------|--------------|--------------|
| Primary | 100 | 25 | 4000 | 6.25 |
| Secondary | 100 | 125 | 800 | 156.25 |
---
# (c) Solve using per-unit
1. **Convert load impedance to per-unit on secondary base:**
\[
Z_{\text{pu}} = \frac{Z_{\text{actual}}}{Z_{b,\text{sec}}} = \frac{30\,\Omega}{156.25\,\Omega} \approx 0.192\,\text{pu}
\]
2. **Secondary voltage in per-unit:**
\[
V_{s,\text{pu}} = \frac{V_s}{V_{b,\text{sec}}} = \frac{125\,\text{kV}}{125\,\text{kV}} = 1.0
\]
3. **Secondary current in per-unit:**
\[
I_{s,\text{pu}} = \frac{V_{s,\text{pu}}}{Z_{\text{pu}}} = \frac{1.0}{0.192} \approx 5.208333\,\text{pu}
\]
Convert back to amps:
\[
I_s = I_{s,\text{pu}} \times I_{b,\text{sec}} = 5.208333 \times 800\,\text{A} \approx 4166.667\,\text{A}
\]
4. **Per-unit load power:**
\[
S_{\text{pu}} = V_{s,\text{pu}} \times I_{s,\text{pu}} = 1.0 \times 5.208333 \approx 5.208333\,\text{pu}
\]
Convert back to actual:
\[
S = 5.208333 \times 100\,\text{MVA} \approx 520.833\,\text{MVA}
\]
5. **Primary current (using turns ratio \(a=5\))**:
\[
I_p = a \times I_s = 5 \times 4166.667\,\text{A} \approx 20{,}833.333\,\text{A} \approx 20.833\,\text{kA}
\]
In primary per-unit (with \(I_{b,\text{primary}}=4000\,\text{A}\)):
\[
I_{p,\text{pu}} = \frac{20{,}833.333}{4000} \approx 5.208333\,\text{pu}
\]
---
### **Final answers (c):**
| Quantity | Value |
|-------------------------|------------------------------|
| Load power | **520.833 MVA** |
| Source current | **20.83 kA** |
*Note:* The per-unit calculations confirm the direct calculation results.
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