A circuit consists of two 475 -k %u03A9 resistors in series with an ideal 12.0-V

Question

A circuit consists of two 475-k%u03A9 resistors in series with an ideal 12.0-V battery.

a) Calculate the potential drop across one of the resistors.

b) A voltmeter with internal resistance 17 M%u03A9 is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

Explanation / Answer

the resistors are R1 = R2 = 475 kohm

they are in series therefore the effective resistance is

R = R1 + R2

in series the current flowing through the resistors is constant therefore

V = I x R

or I = (V/R)

where V = 12.0 V

the potential drop across the resistors is

V1 = I x R1

and V2 = I x R2

the internal resistance of voltmeter is r = 17 Mohm

it is in parallel with R therefore

R' = (1/r) + (1/R)

in parallel the potential difference is constant across the resistors therefore

V = I' x R'

or I' = (V/R')

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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