The ratio of the number of turns of the primary to the secondary coils in an ideal transformer is 20:1. If 240 V AC is applied from a source to the primary coil of the transformer and a 6.0 $ \Omega $ resistor is connected across the output terminals, then the current drawn by the transformer from the source will be:

Question

A coil has $N$ turns and current passes through it is $I$ ampere then we obtain $L$ of self inductance. Now if current is made doubled then new self inductance be ______ H.

Answer 1

To resolve this, we apply transformer equations and Ohm's Law. Given a primary to secondary turns ratio ($N_p:N_s$) of 20:1, an ideal transformer maintains voltage and current ratios proportional to these turns. Consequently, the secondary coil voltage ($V_s$) is calculated as:

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]

With $V_p = 240 \text{V}$ and $N_p:N_s = 20:1$, we determine:

\[ V_s = \frac{V_p \times N_s}{N_p} = \frac{240 \times 1}{20} = 12 \text{V} \]

Next, utilizing Ohm's Law ($V = IR$) and the secondary coil resistance ($R = 6.0 \, \Omega$), we find the secondary coil current ($I_s$):

\[ I_s = \frac{V_s}{R} = \frac{12}{6} = 2 \text{A} \]

An ideal transformer conserves power, meaning primary coil power ($P_p$) equals secondary coil power ($P_s$). Therefore:

\[ P_p = V_p \times I_p = P_s = V_s \times I_s \]

Solving for the primary current ($I_p$):

\[ I_p = \frac{V_s \times I_s}{V_p} = \frac{12 \times 2}{240} = 0.10 \text{A} \]

The current drawn by the transformer from the source is therefore $0.10 \, \text{A}$.

Correct Answer: 0.10 A

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The Correct Option is D

Solution and Explanation

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