hello
i have, learning the art of electronics and the art of electronics books. i am reading the parts about voltage dividers and i am confused about the resistance readings that your supposed to get with meters that measure the outputs and how they load the circuit.
it talks about voltage dividers, for instance a 20v source with two 15k resistors. when the source is seen as zero ohms or ground the resistors are in parallel and half of one resistor which is 7.5k. the output voltage is half the input if both resistors are equal. thevenin equivalent circuit should be the output voltage as source with the parallel resistance as thevenin resistance. i have simulated this in multisim and understand it. but then it goes onto adding a meter which loads the circuit. it has a number of dividers all with the same input and output but different thevenin resistances. then you have to take into consideration the tolerance of the resistor, here it is 1%. the meter is put on the output and a voltage is taken. where the voltage is different than expected it is the loading of the meter that causes it. with a 10v source and 8 volts at the output there is 2 volts across the internal resistance. this should allow for calculation of the internal resistance which should be in proportion to the voltages. 8 volts across the thevenin resistance and 2 volts across the internal resistance, 8 / 2 = 4. so 4 times the thevenin resistance should give you the internal resistance.
how do you get to the point where you can look at the circuit and tell without calculating what the result is? is there a better explanation of this?
i dont like the idea that i have chosen a hard subject to learn, maybe a chose a poor quality book or the example is not clear enough? does anyone else feel that the art of electronics (student manual third edition) is a little tricky?
i have, learning the art of electronics and the art of electronics books. i am reading the parts about voltage dividers and i am confused about the resistance readings that your supposed to get with meters that measure the outputs and how they load the circuit.
it talks about voltage dividers, for instance a 20v source with two 15k resistors. when the source is seen as zero ohms or ground the resistors are in parallel and half of one resistor which is 7.5k. the output voltage is half the input if both resistors are equal. thevenin equivalent circuit should be the output voltage as source with the parallel resistance as thevenin resistance. i have simulated this in multisim and understand it. but then it goes onto adding a meter which loads the circuit. it has a number of dividers all with the same input and output but different thevenin resistances. then you have to take into consideration the tolerance of the resistor, here it is 1%. the meter is put on the output and a voltage is taken. where the voltage is different than expected it is the loading of the meter that causes it. with a 10v source and 8 volts at the output there is 2 volts across the internal resistance. this should allow for calculation of the internal resistance which should be in proportion to the voltages. 8 volts across the thevenin resistance and 2 volts across the internal resistance, 8 / 2 = 4. so 4 times the thevenin resistance should give you the internal resistance.
how do you get to the point where you can look at the circuit and tell without calculating what the result is? is there a better explanation of this?
i dont like the idea that i have chosen a hard subject to learn, maybe a chose a poor quality book or the example is not clear enough? does anyone else feel that the art of electronics (student manual third edition) is a little tricky?