CBSE 2025, Set 2

Question

(b) (i) Write the principle of working of a transformer. With the help of a labelled diagram, explain the working of a step-up transformer.
(ii) An ideal transformer is designed to convert $50	ext{ V}$ into $250	ext{ V}$. It draws $200	ext{ W}$ power from an ac source whose instantaneous voltage is given by $v_i = 20 \sin(100\pi)t	ext{ V}$. Find:
(I) rms value of input current.
(II) expression for instantaneous output voltage.
(III) expression for instantaneous output current.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Transformer:** * **Principle:** It works on the principle of **mutual induction**. When an alternating current flows in the primary coil, it produces a changing magnetic flux that links with the secondary coil, inducing an alternating emf in it. ![Step-Up Transformer](https://www.edvaya.com/images/cbse/physics/2025/set2/q33b-transformer-answer.webp) * **Working (Step-up):** An alternating voltage applied to the primary produces an alternating magnetic flux in the soft iron core. This flux links with the secondary coil. Because a step-up transformer has more turns in the secondary coil than the primary ($N_s > N_p$), and induced emf is proportional to the number of turns ($\epsilon = -N \frac{d\Phi}{dt}$), a higher emf is developed across the secondary windings. **(ii) Transformer Calculations:** Given: Design conversion = $50	ext{ V}$ to $250	ext{ V}$. Therefore, the transformation ratio $\frac{V_o}{V_i} = \frac{250}{50} = 5$. Input power $P_i = 200	ext{ W}$. Instantaneous input voltage $v_i = 20 \sin(100\pi t)	ext{ V}$. Thus, peak input voltage $V_{ip} = 20	ext{ V}$. **(I) RMS value of input current ($I_{p,rms}$):** RMS input voltage $V_{i,rms} = \frac{V_{ip}}{\sqrt{2}} = \frac{20}{\sqrt{2}}	ext{ V}$. Input power $P_i = V_{i,rms} I_{p,rms}$ $$200 = \left(\frac{20}{\sqrt{2}}ight) I_{p,rms}$$ $$I_{p,rms} = \frac{200 \sqrt{2}}{20} = \boxed{10\sqrt{2}	ext{ A}}$$ **(II) Expression for instantaneous output voltage ($v_o$):** Since $\frac{V_o}{V_i} = 5$, the peak output voltage is $V_{op} = 5 	imes V_{ip} = 5 	imes 20 = 100	ext{ V}$. Assuming an ideal transformer, the output voltage is in phase. $$v_o = V_{op} \sin(100\pi t)$$ $$\boxed{v_o = 100 \sin(100\pi t)	ext{ V}}$$ **(III) Expression for instantaneous output current ($i_o$):** For an ideal transformer, Output Power = Input Power = $200	ext{ W}$. $P_o = V_{o,rms} I_{o,rms}$ $$200 = \left(\frac{100}{\sqrt{2}}ight) I_{o,rms}$$ $$I_{o,rms} = \frac{200 \sqrt{2}}{100} = 2\sqrt{2}	ext{ A}$$ Peak output current $I_{op} = I_{o,rms} 	imes \sqrt{2} = (2\sqrt{2}) 	imes \sqrt{2} = 4	ext{ A}$. Assuming a purely resistive load, the current is in phase with the voltage: $$i_o = I_{op} \sin(100\pi t)$$ $$\boxed{i_o = 4 \sin(100\pi t)	ext{ A}}$$

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Physics > Magnetism Induction And Alternating Current > Alternating Current

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