Electromagnetic Induction and Alternating Currents Quiz - ICSE Class 12 Physics
Free practice quiz: Electromagnetic Induction and Alternating Currents (ICSE Class 12 Physics). 44 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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In an AC circuit, the current lags behind the voltage by an angle $\phi$. The power factor of the circuit is given by:
- $\sin \phi$
- $\tan \phi$
- $\cos \phi$
- $1/\cos \phi$
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Answer: C. $\cos \phi$
The power factor in an AC circuit is defined as the cosine of the phase difference between the voltage and the current. It represents the ratio of the real power consumed to the apparent power supplied.
A step-down transformer has a turns ratio of 10:1. If the input voltage is 220 V, what is the output voltage?
- 2200 V
- 22 V
- 2.2 V
- 440 V
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Answer: B. 22 V
A step-down transformer decreases voltage. The turns ratio of 10:1 means $N_p/N_s = 10/1$. For an ideal transformer, $V_p/V_s = N_p/N_s$. So, $220 ext{ V} / V_s = 10/1$. This gives $V_s = 220 ext{ V} / 10 = 22$ V.
In an AC circuit, the impedance $Z$ of a series RL circuit is given by $Z = \sqrt{R^2 + X_L^2}$. If the frequency of the AC source is increased, the impedance will:
- Decrease
- Increase
- Remain the same
- Become zero
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Answer: B. Increase
The inductive reactance $X_L = 2\pi f L$. Since $L$ and $R$ are constant, if the frequency $f$ increases, $X_L$ increases. As $Z = \sqrt{R^2 + X_L^2}$, an increase in $X_L$ will lead to an increase in $Z$.
In a series AC circuit, the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. If $X_L = 50 \Omega$, $X_C = 30 \Omega$, and $R = 40 \Omega$, the phase angle $\phi$ is:
- $\tan^{-1}(1/2)$
- $\tan^{-1}(2/3)$
- $\tan^{-1}(3/4)$
- $\tan^{-1}(4/3)$
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Answer: A. $\tan^{-1}(1/2)$
The phase angle $\phi$ is given by $\tan \phi = (X_L - X_C)/R$. Substituting the given values, $\tan \phi = (50 \Omega - 30 \Omega) / 40 \Omega = 20 \Omega / 40 \Omega = 1/2$. Therefore, $\phi = \tan^{-1}(1/2)$.
A solid iron core is used in a transformer to:
- Increase the resistance
- Decrease the inductance
- Increase the mutual inductance
- Reduce the eddy currents
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Answer: C. Increase the mutual inductance
An iron core has a high magnetic permeability. When placed in a transformer, it concentrates the magnetic field lines, effectively channeling the flux from the primary to the secondary coil. This greatly increases the magnetic flux linkage and thus the mutual inductance between the coils, leading to a more efficient transfer of energy.