A 0.75-M Ohm voltmeter is placed in parallel with a 77.5-k Ohm resistor which is

Question

A 0.75-M Ohm voltmeter is placed in parallel with a 77.5-k Ohm resistor which is in a circuit. (a) What is the resistance, in kilo ohms, of the combination? (b) If the voltage across the combination is kept the same as it was across the 77.5-k Ohm resistor alone, what is the percent increase in the current through this pan of the circuit? (c) If instead the current through the combination is kept the same as it was through the 77.5-k Ohm resistor alone, what is the percentage decrease in voltage?

Explanation / Answer

b)

if voltage is kept same

extra current drawn will be

i = V/0.75x10^6

where as initial current was V/77.5x10^3

percntage increase =100x (V/0.75x10^6)/ V/77.5x10^3

percentage increase = 10.33 %

c)

keeping current constant

so initial voltage will be 77500i

final voltage will be 70240i

percentage decrease in voltage = (77500i- 70240i)100/ 77500i

percentage decrease = 9.36%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.