Homework Solution: The output voltage is given by the voltage divider equation: V_2 = V_1 R_2/R_1 + R_2. Suppose the voltage source produces a volta…

    The output voltage is given by the voltage divider equation: 21 R1+R2 suppose the voltage source produces a voltage of V,-5V with an accuracy of ±10mV, the resistor R1 has a nominal value of 1k(2 with a tolerance of ±5%, and we measure the voltage V2 with our voltmeter to be 3.5V and we know the error in our voltmeter reading to be ±20mV. If we use the above voltage divider equation to calculate the value of the resistor, R2, include the following derivations in your prelab: (1) What is the calculated value of R2? (2) What kind of error should we expect in our calculated value of R? That is, what is the maximum and minimum amount we could expect to be off by?
    The output voltage is given by the voltage divider equation: V_2 = V_1 R_2/R_1 + R_2. Suppose the voltage source produces a voltage of V_1 = 5V with an accuracy of plusminus 10mV, the resistor R_1 has a nominal value of 1k ohm with a tolerance of plusminus 5%, and we measure the voltage V_2 with our voltmeter to be 3.5V and we know the error in our voltmeter reading to be plusminus 20mV. If we use the above voltage divider equation to calculate the value of the resistor, R_2, include the following derivations in your prelab: What is the calculated value of R_2? What kind of error should we expect in our calculated value of R? That is, what is the maximum and minimum amount we could expect to be off by?

    Expert Answer

    The output voltage is loving by the voltage divider equation: 21 R1+R2 imagine the voltage careason produces a voltage of V,-5V with an success of ±10mV, the resistor R1 has a trifling esteem of 1k(2 with a tolerance of ±5%, and we gauge the voltage V2 with our voltmeter to be 3.5V and we recognize the blunder in our voltmeter lection to be ±20mV. If we reason the overhead voltage divider equation to number the esteem of the resistor, R2, enclose the subjoined derivations in your prelab: (1) What is the numberd esteem of R2? (2) What peel of blunder should we foresee in our numberd esteem of R? That is, what is the consummation and reserve aggregate we could foresee to be impromptu by?

    The output voltage is loving by the voltage divider equation: V_2 = V_1 R_2/R_1 + R_2. Imagine the voltage careason produces a voltage of V_1 = 5V with an success of plusminus 10mV, the resistor R_1 has a trifling esteem of 1k ohm with a tolerance of plusminus 5%, and we gauge the voltage V_2 with our voltmeter to be 3.5V and we recognize the blunder in our voltmeter lection to be plusminus 20mV. If we reason the overhead voltage divider equation to number the esteem of the resistor, R_2, enclose the subjoined derivations in your prelab: What is the numberd esteem of R_2? What peel of blunder should we foresee in our numberd esteem of R? That is, what is the consummation and reserve aggregate we could foresee to be impromptu by?

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