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th's magnetic A) 22 μT T-Attenuat Ba anometer has K-0.1A, N-50 and a-5cm, the horizontal comp field Bo is (uo-4mx10 T.m/A): B) 52 µT C) 45 pT D) 39 T EJ 62 µT

th's magnetic A) 22 μT T-Attenuat Ba anometer has K-0.1A, N-50 and a-5cm, the horizontal comp field Bo is (uo-4mx10 T.m/A): B) 52 µT C) 45 pT D) 39 T EJ 62 µT

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the tangent galvanometer has K-0.1A, N-50 and a-5cm, the horizontal compu Earth's magnetic field Bo is (uo-4x10T.m/A): A) 22 µT B) 52 µT C) 45 pT D) 39 T EJ 62 µT T-Attenuator: (Q14, Q15, Q16) 214) If the attenuation factor N-0.2, RL-4000 and e=10 volts, the current passing throu the load resistance is: A) 0.005 A B) 0.004 A C) 0.006 A D) 0.008 A E) 0.02 A Q15) The equivalent resistance in the previous circuit is: A) 5000 B) 2000 C) 2500 D) 4000 E) 1000 Q16) The suggested attenuation factor N if the output voltage across R, becomes 2V is: A) 1.5 B) 3.5 C) 4 D) 0.3 E) 5 The Wheatstone Bridge: (Q17, Q18) 17) The figure shows Wheatstone bridge at equilibrium given that E-SV, R= 200, L₁4 and R2-10, then R. equals A) 40 B) 3.70 C) 8.50 D) 1.20 E) 60 160 Q18) According to the figure in the previous question, if E=5V, then IG equals: www. www.w R Rat 6 Ohms R A) Zero B) 1.5A E) 9A C) 3A D) 6A Electric Power: (Q19, Q20) Q19: A load resistance R, is connected in series to two resistors R₁-120 and R-100 and power supply with E=25V, then the maximum power dissipated in R, in Watts equals: E) 10 B) 2.1 C) 6.4 D) 7.1 A) 5 D120 Q20: If R₁ 200 02, i=0.5A and E-50V, the voltage across R, in Volts when P is maximum El 25 A) 5 C) 15 B) 10
Transcribed Image Text:the tangent galvanometer has K-0.1A, N-50 and a-5cm, the horizontal compu Earth's magnetic field Bo is (uo-4x10T.m/A): A) 22 µT B) 52 µT C) 45 pT D) 39 T EJ 62 µT T-Attenuator: (Q14, Q15, Q16) 214) If the attenuation factor N-0.2, RL-4000 and e=10 volts, the current passing throu the load resistance is: A) 0.005 A B) 0.004 A C) 0.006 A D) 0.008 A E) 0.02 A Q15) The equivalent resistance in the previous circuit is: A) 5000 B) 2000 C) 2500 D) 4000 E) 1000 Q16) The suggested attenuation factor N if the output voltage across R, becomes 2V is: A) 1.5 B) 3.5 C) 4 D) 0.3 E) 5 The Wheatstone Bridge: (Q17, Q18) 17) The figure shows Wheatstone bridge at equilibrium given that E-SV, R= 200, L₁4 and R2-10, then R. equals A) 40 B) 3.70 C) 8.50 D) 1.20 E) 60 160 Q18) According to the figure in the previous question, if E=5V, then IG equals: www. www.w R Rat 6 Ohms R A) Zero B) 1.5A E) 9A C) 3A D) 6A Electric Power: (Q19, Q20) Q19: A load resistance R, is connected in series to two resistors R₁-120 and R-100 and power supply with E=25V, then the maximum power dissipated in R, in Watts equals: E) 10 B) 2.1 C) 6.4 D) 7.1 A) 5 D120 Q20: If R₁ 200 02, i=0.5A and E-50V, the voltage across R, in Volts when P is maximum El 25 A) 5 C) 15 B) 10
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