Electric circuits miscellaneous
-  In the series Rc circuit shown in the figure, the voltage across C starts increasing when the d.c. source is switched on. The rate of increase of voltage across C at the instant just after the switch is closed (i.e. at t = 0+) will be 
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                        View Hint View Answer Discuss in Forum Voltage across the capacitor at any time t, Vc = V (l – e– t / RC) 
 = l – e– t / RC ... (since V = 1 volt)∴ dVc = 1 e– t / RC dt RC At t = 0+ , dVc = 1 dt RC 
 Correct Option: DVoltage across the capacitor at any time t, Vc = V (l – e– t / RC) 
 = l – e– t / RC ... (since V = 1 volt)∴ dVc = 1 e– t / RC dt RC At t = 0+ , dVc = 1 dt RC 
 
-  A periodic rectangular signal X(t) has the wave form shown in the figure. Frequency of the fifth harmonic of its spectrum is 
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                        View Hint View Answer Discuss in Forum Periodic time = 4 ms = 4 × 10 – 3 sec. Fundamental frequency = 103 = 250 Hz 4 
 ∴ Frequency of the 5th harmonic = 250 × 5 = 1250 Hz
 Correct Option: DPeriodic time = 4 ms = 4 × 10 – 3 sec. Fundamental frequency = 103 = 250 Hz 4 
 ∴ Frequency of the 5th harmonic = 250 × 5 = 1250 Hz
 
-  In the given figure, A1, A2 and A3 are ideal ammeters. If A1 reads 5A, A2 reads 12 A, then A3 should read 
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                        View Hint View Answer Discuss in Forum Since the source is an a.c voltage, therefore 
 I 3rms = √I1rms ² + I2rms ²
 = √5² +12² = 13 ACorrect Option: CSince the source is an a.c voltage, therefore 
 I 3rms = √I1rms ² + I2rms ²
 = √5² +12² = 13 A
-  The voltage VC1, VC2 and VC3 across the capacitors in the circuit in the given figure, under steady state are respectively 
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                        View Hint View Answer Discuss in Forum In steady state, capacitors are open and inductances are short.  
 For VC1VC1 = 100 × 40 = 80 V 50 
 For VC2 and VC3VC2 = 80 × C3 C2 + C3 = 80 × 3 = 48 V 5 VC3 = 80 × C2 C2 + C3 
 = 16 × 2 = 32 V
 Correct Option: BIn steady state, capacitors are open and inductances are short.  
 For VC1VC1 = 100 × 40 = 80 V 50 
 For VC2 and VC3VC2 = 80 × C3 C2 + C3 = 80 × 3 = 48 V 5 VC3 = 80 × C2 C2 + C3 
 = 16 × 2 = 32 V
 
-  When the angular frequency ω in the given figure is varied from 0 to ∞ , the locus of the current phasor I2 is given by 
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                        View Hint View Answer Discuss in Forum I2(ω) = Em cosωt R2 + 1 jωC   At ω = ∞ , I2ω = Em R2 
 Also, I2 is leading with voltage phasor.
 At ω = 0, I2 (0) = 0
 Thus desired locus should be as shown.Correct Option: AI2(ω) = Em cosωt R2 + 1 jωC   At ω = ∞ , I2ω = Em R2 
 Also, I2 is leading with voltage phasor.
 At ω = 0, I2 (0) = 0
 Thus desired locus should be as shown.
 
	