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1.25-MΩ voltmeter is placed in parallel with a 74.5-kΩ resistor which is in a circuit.

a. What is the resistance, in kilohms, of the combination?

b. If the voltage across the combination is kept the same as it was across the 74.5-kΩ resistor alone, what is the percent increase in the current through this part of the circuit?

c. If instead the current through the combination is kept the same as it was through the 74.5-kΩ resistor alone, what is the percentage decrease in voltage?

User Vladiki
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1 Answer

6 votes

Final answer:

a. The resistance of the combination is 74.5 kΩ. b. The percent increase in current through the combination is 0%. c. The percentage decrease in voltage across the combination cannot be calculated without the values of voltage and current.

Step-by-step explanation:

a. The resistance of the combination can be calculated using the formula for resistors in parallel, which is given by:

1/R_total = 1/R_1 + 1/R_2 + ... + 1/R_n

where R_total is the total resistance and R_1, R_2, ..., R_n are the resistances of the individual components.

In this case, the combination consists of the 74.5-kΩ resistor and the 1.25-MΩ voltmeter. Plugging in the values:

1/R_total = 1/74.5k + 1/1.25M = 13.4 x 10^-6

Therefore, the resistance of the combination is:

R_total = 74.5kΩ || 1.25MΩ = 1/(13.4 x 10^-6) = 74.5kΩ

b. The current through a resistor is given by Ohm's Law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the voltage across the combination is kept the same as it was across the 74.5-kΩ resistor alone. Therefore, the current through the combination would be:

I_combination = V/R_combination = V/74.5kΩ

The percentage increase in current can be calculated by comparing the current through the combination to the current through the 74.5-kΩ resistor alone:

Percentage increase = ((I_combination - I_resistor alone)/I_resistor alone) x 100

Substituting the values:

Percentage increase = ((V/74.5kΩ - V/74.5kΩ)/V/74.5kΩ) x 100 = 0%

c. To keep the current through the combination the same as it was through the 74.5-kΩ resistor alone, the voltage across the combination needs to change. The voltage across a resistor can be calculated using Ohm's Law: V = I*R, where V is the voltage, I is the current, and R is the resistance. In this case, the current through the combination is kept the same as it was through the 74.5-kΩ resistor alone. Therefore, the voltage across the combination would be:

V_combination = I_resistor alone x R_combination = I_resistor alone x 74.5kΩ

The percentage decrease in voltage can be calculated by comparing the voltage across the combination to the voltage across the 74.5-kΩ resistor alone:

Percentage decrease = ((V_resistor alone - V_combination)/V_resistor alone) x 100

Substituting the values:

Percentage decrease = ((V_resistor alone - I_resistor alone x 74.5kΩ)/V_resistor alone) x 100

However, the values for the voltage and current are not given in the question, so it is not possible to calculate the percentage decrease in voltage.

User Pejmanjohn
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