Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistance shown in shown in figure. The output voltage (V0) is :
- 15 V
- 7 V
- 44 V
- 22 V
The Correct Option is B
Approach Solution - 1
The problem asks to determine the output voltage (\(V_0\)) in the given circuit, which consists of a transformer followed by a resistive voltage divider.
Concept Used:
The solution involves two main concepts:
1. The Transformer Equation: For an ideal transformer, the ratio of the secondary voltage (\(V_s\)) to the primary voltage (\(V_p\)) is equal to the ratio of the number of turns in the secondary coil (\(N_s\)) to the number of turns in the primary coil (\(N_p\)).
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]2. The Voltage Divider Rule: In a series circuit, the voltage across a particular resistor is proportional to its resistance. The voltage (\(V_{out}\)) across a resistor \(R_{out}\) in a series combination with a total resistance \(R_{total}\) and a total input voltage \(V_{in}\) is given by:
\[ V_{out} = V_{in} \times \frac{R_{out}}{R_{total}} \]Step-by-Step Solution:
Step 1: Calculate the voltage across the secondary coil of the transformer (\(V_s\)).
The given values for the transformer are:
- Primary voltage, \(V_p = 220 \, \text{V}\)
- Number of turns in the primary coil, \(N_p = 100\)
- Number of turns in the secondary coil, \(N_s = 10\)
Using the transformer equation:
\[ \frac{V_s}{220 \, \text{V}} = \frac{10}{100} \] \[ V_s = 220 \times \frac{1}{10} = 22 \, \text{V} \]This secondary voltage \(V_s = 22 \, \text{V}\) is the input voltage to the resistive load circuit.
Step 2: Analyze the secondary circuit as a voltage divider.
The secondary coil is connected to two resistors in series:
- \(R_1 = 15 \, \text{k}\Omega\)
- \(R_2 = 7 \, \text{k}\Omega\)
The total resistance of the secondary circuit is:
\[ R_{total} = R_1 + R_2 = 15 \, \text{k}\Omega + 7 \, \text{k}\Omega = 22 \, \text{k}\Omega \]The output voltage \(V_0\) is the voltage across the \(7 \, \text{k}\Omega\) resistor. So, \(R_{out} = 7 \, \text{k}\Omega\).
Step 3: Apply the voltage divider rule to find the output voltage \(V_0\).
The input voltage to this divider circuit is \(V_{in} = V_s = 22 \, \text{V}\).
\[ V_0 = V_{in} \times \frac{R_{out}}{R_{total}} \] \[ V_0 = 22 \, \text{V} \times \frac{7 \, \text{k}\Omega}{22 \, \text{k}\Omega} \]Final Computation & Result:
Simplify the expression to find the final value of \(V_0\).
\[ V_0 = 22 \times \frac{7}{22} \] \[ V_0 = 7 \, \text{V} \]The output voltage (V0) is 7 V.
Approach Solution -2
Calculate the Secondary Voltage Using the Turns Ratio: The turns ratio for a transformer is given by:
\(\frac{\epsilon_1}{\epsilon_2} = \frac{N_1}{N_2}\)
Substitute \(N_1 = 100\), \(N_2 = 10\), and \(\epsilon_1 = 220 \, V\):
\(\epsilon_2 = \frac{N_2}{N_1} \times \epsilon_1 = \frac{10}{100} \times 220 = 22 \, V\)
Determine the Equivalent Resistance of the Load: The load consists of two resistances, \(15 \, \Omega\) and \(7 \, \Omega\), connected in series:
\(R_{eq} = 15 + 7 = 22 \, \Omega\)
Calculate the Current in the Secondary Circuit: Using Ohm’s law for the secondary circuit:
\(I = \frac{\epsilon_2}{R_{eq}} = \frac{22 \, V}{22 \, \Omega} = 1 \, A\)
Calculate the Output Voltage Across the 7 \( \Omega \) Resistor: The output voltage \(V_0\) across the \(7 \, \Omega\) resistor is:
\(V_0 = I \times 7 = 1 \, A \times 7 \, \Omega = 7 \, V\).
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Concepts Used:
Resistance
Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.
R=V/I
In this case,
v = Voltage across its ends
I = Current flowing through it
All materials resist current flow to some degree. They fall into one of two broad categories:
- Conductors: Materials that offer very little resistance where electrons can move easily. Examples: silver, copper, gold and aluminum.
- Insulators: Materials that present high resistance and restrict the flow of electrons. Examples: Rubber, paper, glass, wood and plastic.
Resistance measurements are normally taken to indicate the condition of a component or a circuit.
- The higher the resistance, the lower the current flow. If abnormally high, one possible cause (among many) could be damaged conductors due to burning or corrosion. All conductors give off some degree of heat, so overheating is an issue often associated with resistance.
- The lower the resistance, the higher the current flow. Possible causes: insulators damaged by moisture or overheating.