A power transmission line feeds input power at 2.3 mathrm~kV$2.3 \mathrm{~kV}$ to a step down transformer with its primary winding having 3000$3000$ turns. The output power is delivered at 230 mathrm~V$230 \mathrm{~V}$ by the transformer. The current in the primary of the transformer is 5 mathrmA$5 \mathrm{A}$ and its efficiency is 90\%$90\%$. The winding of transformer is made of copper. The output current of transformer is ________ mathrmA$\mathrm{A}$.
Numerical Answer Type:
Enter a numerical valueAnswer: 45 to 45
Solution & Explanation
### Related Formula
eta = fracP_textoutP_textin$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$P_textin = V_p I_p$P_{\text{in}} = V_p I_p$P_textout = V_s I_s$P_{\text{out}} = V_s I_s$
### Core Logic
The efficiency eta$\eta$ of a transformer is the ratio of output power to input power. We can use this to find the secondary (output) current.
### Step 1: Calculate Input Power
Given:
V_p = 2.3 mathrm~kV = 2300 mathrm~V$V_p = 2.3 \mathrm{~kV} = 2300 \mathrm{~V}$I_p = 5 mathrm~A$I_p = 5 \mathrm{~A}$P_textin = 2300 times 5 mathrm~W$P_{\text{in}} = 2300 \times 5 \mathrm{~W}$
### Step 2: Calculate Output Power and Current
Efficiency eta = 90\% = 0.9$\eta = 90\% = 0.9$P_textout = eta times P_textin = 0.9 times 2300 times 5$P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 2300 \times 5$
Since P_textout = V_s I_s$P_{\text{out}} = V_s I_s$ and V_s = 230 mathrm~V$V_s = 230 \mathrm{~V}$:
V_s I_s = 230 times I_s = 0.9 times 2300 times 5$V_s I_s = 230 \times I_s = 0.9 \times 2300 \times 5$I_s = frac0.9 times 2300 times 5230 = 0.9 times 10 times 5$I_s = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5$I_s = 9 times 5 = 45 mathrm~A$I_s = 9 \times 5 = 45 \mathrm{~A}$
### Pattern Recognition
Transformer equations are direct: eta V_p I_p = V_s I_s$\eta V_p I_p = V_s I_s$. The number of turns (3000$3000$) is distractor data not needed unless calculating the secondary turns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Keywords:#output current of transformer#JEE Main 2024 Evening Q51#Alternating Current JEE Main 2024#Transformer Efficiency JEE Main 2024
More Electromagnetic Induction Previous-Year Questions — Page 4
Q57jee_main_2024_31_jan_morningMutual Inductance
A small square loop of wire of side ell$\ell$ is placed inside a large square loop of wire of side L$L$ (L = ell^2$L = \ell^2$). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is sqrtx times 10^-7mathrm\ H$\sqrt{x} \times 10^{-7}\mathrm{\ H}$, where x =$x =$
Numerical Answer.Answer: 128 to 128
Solution
### Related Formula
M = fracphi_2i_1$M = \frac{\phi_2}{i_1}$B_textstraight wire segment = fracmu_0 i4pi d (sintheta_1 + sintheta_2)$B_{\text{straight wire segment}} = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$
### Core Logic
Mutual Inductance diagram for Q57 - JEE Main 2024 Morning
Assume a current i$i$ flows through the larger square loop of side L$L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side ell$\ell$.
The magnetic field at the center of the large square loop (distance d = L/2$d = L/2$ from each side, angles 45^circ$45^\circ$):
B = 4 times left[ fracmu_0 i4pi (L/2) (sin 45^circ + sin 45^circ) right]$B = 4 \times \left[ \frac{\mu_0 i}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \right]$B = fracmu_0 ipi (L/2) left( frac2sqrt2 right)$B = \frac{\mu_0 i}{\pi (L/2)} \left( \frac{2}{\sqrt{2}} \right)$B = frac2sqrt2 mu_0 ipi L$B = \frac{2\sqrt{2} \mu_0 i}{\pi L}$
### Step 2: Mutual Inductance Calculation
Flux linkage for the inner loop:
phi = B cdot ell^2$\phi = B \cdot \ell^2$phi = frac2sqrt2 mu_0 ipi L ell^2$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$
Given L = ell^2$L = \ell^2$:
phi = frac2sqrt2 mu_0 ipi (ell^2) ell^2 = frac2sqrt2 mu_0 ipi$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi (\ell^2)} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$
Mutual inductance M$M$:
M = fracphii = frac2sqrt2 mu_0pi$M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0}{\pi}$
Using mu_0 = 4pi times 10^-7$\mu_0 = 4\pi \times 10^{-7}$:
M = frac2sqrt2 times 4pi times 10^-7pi$M = \frac{2\sqrt{2} \times 4\pi \times 10^{-7}}{\pi}$M = 8sqrt2 times 10^-7mathrm\,H$M = 8\sqrt{2} \times 10^{-7}\mathrm{\,H}$M = sqrt128 times 10^-7mathrm\,H$M = \sqrt{128} \times 10^{-7}\mathrm{\,H}$
Comparing with sqrtx times 10^-7$\sqrt{x} \times 10^{-7}$, we get x = 128$x = 128$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
More Electromagnetic Induction Questions — jee_main_2024_30_january_evening
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