In an ideal transformer, voltage and the current in the primary coil are 200V and 2A, respectively. If the voltage in the secondary coil is 2000 V, then the value of current in the secondary coil will be
$\begin{align}
& A.0.2A \\
& B.2A \\
& C.10A \\
& D.20A \\
\end{align}$
Answer
672k+ views
Hint: Transformer is a passive electrical device that transfers electrical energy from one circuit to another circuit through the process of electromagnetic induction without any change in the input power and output power.
Step by step solution:
We are given the primary voltage is 200 volts and the primary current is given as 2 Amperes. The secondary cell voltage is given as 2000 volts.
Let us take secondary coil current as $I_s$ amperes
We know that the transformer transfers electrical energy from one circuit to another circuit without any change in the power at the input and output terminals.
We know the power is given by power = voltage x current
So the power at the input terminals becomes $P_{input} = V_p \times I_p = 200 \times 2 = 400W$
The power at the output terminals becomes $P_{output} = V_s \times I_s = 2000 \times I_s = 2000I_s W$
As the input power is equal to output power we get
$P_{input} = P_{output}$
So now the unknown secondary coil current can be calculated as:
$2000I_s = 400$
$I_s = \dfrac{400}{2000} = 0.2 A$
Thus, we have calculated the unknown secondary coil current using the concept of power transfer in ideal transformer and we got the secondary current value as $I_s = 0.2 A$
Note: The given problem can be solved in another way using the relation between voltage and number of turns and relation between current and number of turns in the coil
using the two relations. We have $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$and $\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$ . we can show that the product of voltage and current on both sides' primary and secondary sides of the transformer are equal which we used in the above problem. So instead of using the number of the turns ratio, we can directly use the concept of power in the above problem.
Step by step solution:
We are given the primary voltage is 200 volts and the primary current is given as 2 Amperes. The secondary cell voltage is given as 2000 volts.
Let us take secondary coil current as $I_s$ amperes
We know that the transformer transfers electrical energy from one circuit to another circuit without any change in the power at the input and output terminals.
We know the power is given by power = voltage x current
So the power at the input terminals becomes $P_{input} = V_p \times I_p = 200 \times 2 = 400W$
The power at the output terminals becomes $P_{output} = V_s \times I_s = 2000 \times I_s = 2000I_s W$
As the input power is equal to output power we get
$P_{input} = P_{output}$
So now the unknown secondary coil current can be calculated as:
$2000I_s = 400$
$I_s = \dfrac{400}{2000} = 0.2 A$
Thus, we have calculated the unknown secondary coil current using the concept of power transfer in ideal transformer and we got the secondary current value as $I_s = 0.2 A$
Note: The given problem can be solved in another way using the relation between voltage and number of turns and relation between current and number of turns in the coil
using the two relations. We have $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$and $\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$ . we can show that the product of voltage and current on both sides' primary and secondary sides of the transformer are equal which we used in the above problem. So instead of using the number of the turns ratio, we can directly use the concept of power in the above problem.
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