Đề song ngữ Vật lí 9 - Đề 5 - Trường THCS Đông Tây Hưng (Có đáp án)

 ĐỀ 5
Question 1: To transmit the power capacity 100kW from the electric wire which hasthe 
resistance5Ω with waste capacity 0,5kW. The potential difference between 2 electric wires is ..
A. 10kV B. 15kV C. 20kV D. 5kV
Question 2:Given 3 resistances R1 = 10Ω; R2 = R3 = 5Ω. Connect the electric circuits [R1 is parallel 
with (R2 is serial R3))] then connect 6V – potential difference power source. The amperage of the 
main circuit is .
A. 1,2A B. 0,2A C. 0,6A D. 0,3A
Question 3 :Give the circuit [R1 is serial (R2 parallels R3)] , R1 = 3Ω; R2 = R3 = 2R1 andThe 
amperage of the main circuit is1500mA. The potential difference between 2 circuits is........A.6V
 B. 9V C. 4,5V D. 3V
Question 4:A iron operates continuously in 2 hours.The potential difference is 220V.The indices of 
electric meter increases 2 .The power of the iron is .
A. 1000W B. 2000W C. 100W D. 500W
Question 5:Put the potential difference12V in the circuit, the electricity is used to in 30 minutes is 
43,2kJ.Put the potential difference 15V in the circuit,,the electricity is used to in 1 hour is ..A.
 135kJ B. 1350kJ C. 1,35kJ D. 13,5kJ
Question 6:Put a constant resistance 6V in two side of the electric wire then the amperage through 
that electric wire is 50mA. The resistance of the electric wire is . Ω ?
Question 7: Give resistances R1; R2; R3; are in series and connectto constant 12V - potential 
difference power source. R1 = 1,5R2 = 3R3, potential difference between 2 resistances R3 is 
 V
Question 8: Question 9: chưa học
Question 10: Two lights Đ1 (6V - 6W), Đ2 (6V - 3W) are bright normal. The current ratio of I1: I2 
runs through two filament lamps above ...............:
Question 11: The circuit consists of two resistors R1 = 12 , R2 = 6  parallel connection . On two 
points there is voltage Potential difference 12V
 a. Calculate the equivalent resistance of the circuit.
 b. Calculate the amperage through each resistor and through the main circuit.
 c. Calculate the heat dissipated on the circuit for 10 minutes. Answer
Key:
 Question 1 2 3 4 5 6 7 8 9 10
 Key 120 2
 A A B A A 2: 1
 (Ω) (V)
 Answer 11
 a.4
 b. 2 A
 c. 21600 J The amperage of the main circuit is: 
 Because: R1 parallel R2 so U=U1=U2
 The intensity of electric current passing through resistor R1 is: 
 The intensity of electric current passing through resistor R2 is:
d. the heat dissipated on the circuit for 10 minutes is:
 Q = I2.R.t = 32.4.(10.60) = 21600 (J)