AC Generator and Transformer — JEE Main Physics MCQs with Solutions
Free JEE Main Physics AC Generator and Transformer MCQs with step-by-step solutions (9 questions). Part of Alternating Current. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — AC Generator and Transformer · medium · numerical
A step down transformer connected to an AC mains supply of $220\,\text{V}$ is made to operate at $11\,\text{V}$, $44\,\text{W}$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
A. $0.2\,\text{A}$ ✓ Correct
B. $0.4\,\text{A}$
C. $2\,\text{A}$
D. $4\,\text{A}$
Solution: Since power losses are ignored, power in primary equals power in secondary: $P = V_P I_P$. So $44 = 220 \times I_P$, giving $I_P = \dfrac{44}{220} = 0.2\,\text{A}$.
Q2 — AC Generator and Transformer · medium · theory
The core of a transformer is laminated because
A. energy losses due to eddy currents may be minimised ✓ Correct
B. the weight of the transformer may be reduced
C. rusting of the core may be prevented
D. ratio of voltage in primary and secondary may be increased
Solution: A changing magnetic flux induces eddy currents in the solid core, wasting energy as heat. Laminating the core breaks up these current loops into smaller loops, increasing resistance and minimising eddy current losses.
Q3 — AC Generator and Transformer · medium · numerical
The primary winding of a transformer has $500$ turns whereas its secondary has $5000$ turns. The primary is connected to an AC supply of $20\,\text{V}$-$50\,\text{Hz}$. The secondary will have an output of
A. $2\,\text{V}, 5\,\text{Hz}$
B. $200\,\text{V}, 500\,\text{Hz}$
C. $2\,\text{V}, 50\,\text{Hz}$
D. $200\,\text{V}, 50\,\text{Hz}$ ✓ Correct
Solution: A transformer changes AC voltage but never changes frequency. Using $V_s = \dfrac{N_s}{N_p} V_p = \dfrac{5000}{500} \times 20 = 200\,\text{V}$, with frequency unchanged at $50\,\text{Hz}$.
Q4 — AC Generator and Transformer · medium · numerical
A transformer having efficiency of $90\%$ is working on $200\,\text{V}$ and $3\,\text{kW}$ power supply. If the current in the secondary coil is $6\,\text{A}$, the voltage across the secondary coil and the current in the primary coil respectively are
A. $300\,\text{V}, 15\,\text{A}$
B. $450\,\text{V}, 15\,\text{A}$ ✓ Correct
C. $450\,\text{V}, 13.5\,\text{A}$
D. $600\,\text{V}, 15\,\text{A}$
Solution: Output power $= 0.90 \times 3000 = 2700\,\text{W}$. Since $V_2 I_2 = 2700\,\text{W}$ and $I_2 = 6\,\text{A}$, $V_2 = \dfrac{2700}{6} = 450\,\text{V}$. Input power: $V_1 I_1 = 3000\,\text{W}$, with $V_1 = 200\,\text{V}$, so $I_1 = \dfrac{3000}{200} = 15\,\text{A}$.
Q5 — AC Generator and Transformer · medium · numerical
A $220\,\text{V}$ input is supplied to a transformer. The output circuit draws a current of $2.0\,\text{A}$ at $440\,\text{V}$. If the efficiency of the transformer is $80\%$, the current drawn by the primary windings of the transformer is
A. $3.6\,\text{A}$
B. $2.8\,\text{A}$
C. $2.5\,\text{A}$
D. $5.0\,\text{A}$ ✓ Correct
Solution: Efficiency $\eta\% = \dfrac{P_{out}}{P_{input}} \times 100 = \dfrac{V_s i_s}{V_p i_p} \times 100$. So $80 = \dfrac{2 \times 440}{220 \times i_p} \times 100$, giving $i_p = 5\,\text{A}$.
Q6 — AC Generator and Transformer · medium · theory
In an AC circuit the emf ($V$) and the current ($i$) at any instant are given respectively by $V = V_0 \sin \omega t$, $i = i_0 \sin(\omega t - \phi)$. The average power in the circuit over one cycle of AC is
A. $\dfrac{V_0 i_0}{2}$
B. $\dfrac{V_0 i_0}{2} \sin \phi$
C. $\dfrac{V_0 i_0}{2} \cos \phi$ ✓ Correct
D. $V_0 i_0$
Solution: $P_{av} = V_{rms} i_{rms} \cos\phi = \dfrac{1}{2} V_0 i_0 \cos\phi$, since $V_{rms} = \dfrac{V_0}{\sqrt{2}}$ and $i_{rms} = \dfrac{i_0}{\sqrt{2}}$, where $\cos\phi$ is the power factor.
Q7 — AC Generator and Transformer · medium · numerical
A transformer is used to light a $100\,\text{W}$ and $110\,\text{V}$ lamp from a $220\,\text{V}$ mains. If the main current is $0.5\,\text{A}$, the efficiency of the transformer is approximately
A. $30\%$
B. $50\%$
C. $90\%$ ✓ Correct
D. $10\%$
Solution: Efficiency $\eta = \dfrac{\text{Output power}}{\text{Input power}} = \dfrac{V_s i_s}{V_p i_p}$. Given $V_s i_s = 100\,\text{W}$, $V_p = 220\,\text{V}$, $i_p = 0.5\,\text{A}$. So $\eta = \dfrac{100}{220 \times 0.5} \approx 0.90 = 90\%$.
Q8 — AC Generator and Transformer · medium · numerical
The primary and secondary coils of a transformer have $50$ and $1500$ turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi = \phi_0 + 4t$, where $\phi$ is in weber, $t$ is time in second and $\phi_0$ is a constant, the output voltage across the secondary coil is
A. $90\,\text{V}$
B. $120\,\text{V}$ ✓ Correct
C. $220\,\text{V}$
D. $30\,\text{V}$
Solution: Voltage across primary: $V_p = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(\phi_0 + 4t) = 4\,\text{V}$. Since $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$, $V_s = V_p \dfrac{N_s}{N_p} = 4 \times \dfrac{1500}{50} = 120\,\text{V}$.
Q9 — AC Generator and Transformer · medium · numerical
A step-up transformer operates on a $230\,\text{V}$ line and supplies current of $2\,\text{A}$ to a load. The ratio of the primary and secondary windings is $1:25$. The current in the primary coil is
A. $15\,\text{A}$
B. $50\,\text{A}$ ✓ Correct
C. $25\,\text{A}$
D. $12.5\,\text{A}$
Solution: For no loss of power, $V i = \text{constant}$, so $\dfrac{i_p}{i_s} = \dfrac{N_s}{N_p}$. Here $\dfrac{N_p}{N_s} = \dfrac{1}{25}$, so $\dfrac{i_p}{2} = \dfrac{25}{1}$, giving $i_p = 25 \times 2 = 50\,\text{A}$.