VIPSolutionsGive step-by-step solution with explanation and final answer:When a transformer operates under no-load conditions, the secondary winding is open and no current flows through it. In this case, the primary winding draws a small current from the supply, known as no-load current. This no-load current produces a flux in the transformer’s core. As a result, a voltage is induced in the secondary winding, termed as the secondary voltage. The understanding of these quantities is important for grasping the basic principles of transformer operation. Given: 1. Supply voltage V;= 230 V, Secondary turns N,=1200 2. Formula for no-load current: I= n I i= Le 27tf L, ST eR ee i ela current [p= 230 ve VN, 4. Expression 27x50x0,5 M
Question:
Give step-by-step solution with explanation and final answer:
When a transformer operates under
no-load conditions, the secondary
winding is open and no current flows
through it. In this case, the primary
winding draws a small current from
the supply, known as no-load current.
This no-load current produces a flux
in the transformer’s core. As a result,
a voltage is induced in the secondary
winding, termed as the secondary
voltage. The understanding of these
quantities is important for grasping
the basic principles of transformer
operation.
Given:
1. Supply voltage V;= 230 V, Secondary
turns N,=1200
2. Formula for no-load current: I= n
I i= Le 27tf L,
ST eR ee i ela
current [p= 230 ve VN,
4. Expression 27x50x0,5 M
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Asked by: Aman
Created at: 2025-09-30 07:48:15
Asked by: Aman
Created at: 2025-09-30 07:48:15
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Let's break down and solve the problem step-by-step using the information and formulas provided in the image.
---
## **Step 1: List Given Data**
- **Supply voltage** \( V_1 = 230 \) V
- **Secondary turns** \( N_2 = 120 \)
- **Frequency** \( f = 50 \) Hz
- **Inductance** \( L = .5 \) H (from the substitution step)
---
## **Step 2: No-Load Current Formula**
The formula for no-load current is:
\[
I_ = \frac{V_1}{2\pi f L_1}
\]
Given values:
- \( V_1 = 230 \) V
- \( f = 50 \) Hz
- \( L_1 = .5 \) H
---
## **Step 3: Substitute Values**
\[
I_ = \frac{230}{2\pi \times 50 \times .5}
\]
---
## **Step 4: Calculate Denominator**
\[
2\pi \times 50 \times .5 = 2 \times 3.1416 \times 50 \times .5
\]
\[
= 6.2832 \times 50 \times .5
\]
\[
= 314.16 \times .5
\]
\[
= 157.08
\]
---
## **Step 5: Calculate No-Load Current**
\[
I_ = \frac{230}{157.08} \approx 1.464 \text{ A}
\]
---
## **Step 6: Final Answer**
### **No-load current \( I_ \) is:**
\[
\boxed{1.46\ \text{A}}
\]
---
### **Summary Table**
| Quantity | Value |
|----------------------|-----------|
| Supply Voltage (\(V_1\)) | 230 V |
| Frequency (\(f\)) | 50 Hz |
| Inductance (\(L_1\)) | .5 H |
| No-load Current (\(I_\)) | **1.46 A**|
---
If you need to calculate the secondary voltage (\(V_2\)), please provide the value for primary turns (\(N_1\)), as it is missing here.
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