標題:
電學問題 20分
發問:
9. Three resistors of 10 ohm , 15 ohm , and 20 ohm respectively are connected in parallel. This combination is connected in series with another parallel combination of 8 ohm and 5 ohm. If the complete circuit is supplied from a 20V source, calculate the total resistance, the total current,the pd across the 8 ohm... 顯示更多 9. Three resistors of 10 ohm , 15 ohm , and 20 ohm respectively are connected in parallel. This combination is connected in series with another parallel combination of 8 ohm and 5 ohm. If the complete circuit is supplied from a 20V source, calculate the total resistance, the total current,the pd across the 8 ohm resistor and the current through the 5 ohm resistor. 11. Three loads, of 22A,11A and 5.5A are supplied from a 220V source. If a motor of resistance 24 Ohm is also connected across the supply, calculate the total resistance and the total current drawn from the supply. 13. A 220V, 0.5A lamp is to be connected in series with a resistor across a 240V supply. Determine the resistor value required for the lamp to operate as its correct voltage. 15. An electric cooker element consists of two parts,each having a resistor of 15 Ohm,which can be connected in series , in parallel,or using one part inly. Calculate the current drawn from a 220v supply for each connection. 17.A battery has a terminal voltage of 1.8V when supply a current of 10A. This voltage rises to 2.0V when the load is removed. Calculate the internal resistance. 19. Calculate the p.d across the 2 resistor shown in fig Q19 21. A circuit consists of a 15 Ohm and a20 Ohm resistor connected in parallel across a battery of internal resistance 2 Ohm. If 60W is dissipated by the 15 Ohm resistor, calculate the current in the 20 Ohm resistor, the terminal p.d. and the emf of the battery , the total power dissipated in the external circuit. 23. The lamps in a set of Christmas tree lights areconnected in series ; if there are 20 lamps and each lamp has a resistance of 25 Ohm ,calculate the total resistance of the set of lamps , and hence calculate the current taken from a 220V supply.
最佳解答:
9. Total resistance, R = (1/10 + 1/15 + 1/20)-1 + (1/8 + 1/5)-1 = 7.69 ohms By V = IR 20 = I(7.69) Total current, I = 2.6 A P.D. across the 8-ohm resistor = (1/5 + 1/8)-1 / 7.69 X 20 = 8 V P.D. across the 5-ohm resistor = 8 V By V = IR 8 = I(5), I = 1.6 A Current through the 5 ohm resistor = 1.6 A 11. Resistance of the three resistors 220/22 and 220/11 and 220/5.5 respectively = 10 ohms, 20 ohms and 40 ohms Total resistance, R = 10 + 20 + 40 + 24 = 94 ohms Total current, I = V/R = 220 / 94 = 2.34 A 13. For the lamp to operate as its correct voltage, the current must be 0.5 A Hence, current through the resistor, I = 0.5 A Resistance of the resistor, R = V/I = 240/0.5 = 480 ohms 15. In series, resistance, R = 15 + 15 = 30 ohms Current drawn, I = V/R = 220 / 30 = 7.33 A In parallel, resistance, R = (1/15 + 1/15)-1 = 7.5 ohms Current drawn, I = V/R = 220 / 7.5 = 29.3 A One part only: current drawn, I = V/R = 220/15 = 14.7 A 17. From the question, when the current is 10 A, the battery has a potential drop of = 2.0 - 1.8 = 0.2 V Hence, internal resistance, R = V/I = 0.2/10 = 0.02 ohms 19. There is no graph showing your question. 21. Current through the 15 ohm resistor, I' = (60/15)1/2 = 2 A Since the 15 ohm resistor and the 20 ohm resistor are connected in parallel, so the P.D. across both of them are the same. Hence, current through the 20 ohm resistor, I = 2 X 15/20 = 1.5 A Terminal P.D. = P.D. across the resistors = 1.5 X 20 = 30 V Current in the circuit, I = 1.5 + 2 = 3.5 A Equivalent resistance in the circuit, R = (1/15 + 1/20)-1 + 2 = 10.6 ohms E.m.f. of the battery, V = IR = 3.5 X 10.6 = 37 V Power dissipated by the 20 ohm resistor = (1.5)2(20) = 45 W Total power dissipated = 45 + 60 = 105 W 23. Total resistance, R = 20 X 25 = 500 ohms Current, I = V/R = 220 / 500 = 0.44 A
11條個度唔係 in parallel咩??
電學問題 20分
發問:
9. Three resistors of 10 ohm , 15 ohm , and 20 ohm respectively are connected in parallel. This combination is connected in series with another parallel combination of 8 ohm and 5 ohm. If the complete circuit is supplied from a 20V source, calculate the total resistance, the total current,the pd across the 8 ohm... 顯示更多 9. Three resistors of 10 ohm , 15 ohm , and 20 ohm respectively are connected in parallel. This combination is connected in series with another parallel combination of 8 ohm and 5 ohm. If the complete circuit is supplied from a 20V source, calculate the total resistance, the total current,the pd across the 8 ohm resistor and the current through the 5 ohm resistor. 11. Three loads, of 22A,11A and 5.5A are supplied from a 220V source. If a motor of resistance 24 Ohm is also connected across the supply, calculate the total resistance and the total current drawn from the supply. 13. A 220V, 0.5A lamp is to be connected in series with a resistor across a 240V supply. Determine the resistor value required for the lamp to operate as its correct voltage. 15. An electric cooker element consists of two parts,each having a resistor of 15 Ohm,which can be connected in series , in parallel,or using one part inly. Calculate the current drawn from a 220v supply for each connection. 17.A battery has a terminal voltage of 1.8V when supply a current of 10A. This voltage rises to 2.0V when the load is removed. Calculate the internal resistance. 19. Calculate the p.d across the 2 resistor shown in fig Q19 21. A circuit consists of a 15 Ohm and a20 Ohm resistor connected in parallel across a battery of internal resistance 2 Ohm. If 60W is dissipated by the 15 Ohm resistor, calculate the current in the 20 Ohm resistor, the terminal p.d. and the emf of the battery , the total power dissipated in the external circuit. 23. The lamps in a set of Christmas tree lights areconnected in series ; if there are 20 lamps and each lamp has a resistance of 25 Ohm ,calculate the total resistance of the set of lamps , and hence calculate the current taken from a 220V supply.
最佳解答:
9. Total resistance, R = (1/10 + 1/15 + 1/20)-1 + (1/8 + 1/5)-1 = 7.69 ohms By V = IR 20 = I(7.69) Total current, I = 2.6 A P.D. across the 8-ohm resistor = (1/5 + 1/8)-1 / 7.69 X 20 = 8 V P.D. across the 5-ohm resistor = 8 V By V = IR 8 = I(5), I = 1.6 A Current through the 5 ohm resistor = 1.6 A 11. Resistance of the three resistors 220/22 and 220/11 and 220/5.5 respectively = 10 ohms, 20 ohms and 40 ohms Total resistance, R = 10 + 20 + 40 + 24 = 94 ohms Total current, I = V/R = 220 / 94 = 2.34 A 13. For the lamp to operate as its correct voltage, the current must be 0.5 A Hence, current through the resistor, I = 0.5 A Resistance of the resistor, R = V/I = 240/0.5 = 480 ohms 15. In series, resistance, R = 15 + 15 = 30 ohms Current drawn, I = V/R = 220 / 30 = 7.33 A In parallel, resistance, R = (1/15 + 1/15)-1 = 7.5 ohms Current drawn, I = V/R = 220 / 7.5 = 29.3 A One part only: current drawn, I = V/R = 220/15 = 14.7 A 17. From the question, when the current is 10 A, the battery has a potential drop of = 2.0 - 1.8 = 0.2 V Hence, internal resistance, R = V/I = 0.2/10 = 0.02 ohms 19. There is no graph showing your question. 21. Current through the 15 ohm resistor, I' = (60/15)1/2 = 2 A Since the 15 ohm resistor and the 20 ohm resistor are connected in parallel, so the P.D. across both of them are the same. Hence, current through the 20 ohm resistor, I = 2 X 15/20 = 1.5 A Terminal P.D. = P.D. across the resistors = 1.5 X 20 = 30 V Current in the circuit, I = 1.5 + 2 = 3.5 A Equivalent resistance in the circuit, R = (1/15 + 1/20)-1 + 2 = 10.6 ohms E.m.f. of the battery, V = IR = 3.5 X 10.6 = 37 V Power dissipated by the 20 ohm resistor = (1.5)2(20) = 45 W Total power dissipated = 45 + 60 = 105 W 23. Total resistance, R = 20 X 25 = 500 ohms Current, I = V/R = 220 / 500 = 0.44 A
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