Who can solve this interesting physics problem???????

Question

The voltage across the 30 k-Omega resistor in the figure (Intro 1 figure) is measured with (a) a 150 k-Omega voltmeter, (b) a 350 k-Omega voltmeter, and (c) a digital meter with 50 M-Omega resistance. To two significant figures, what does each read?

I cannot copy and paste the figure, its like the 30ko resistor is in series with one 40 ko resistor while this 40 ko resistor is in parallel with another 40 kor esistor.

i have no idea how to do it. can anyone help me with this?

Thank you.

Answer 1

The two 40 k-ohm resistors in parallel act as a 20 k ohm resistor.

When the meters are put across the 30 k-ohm resistor they are in parallel with it.

(a) a 30 k resistor and a 150 k resistor in parallel have an effective resistance of:

Rt = 30*150/(30+150) = 25

Then effectively there are a 25 k resistor in series with a 20 k resistor.

This divides the applied voltage by a ratio of 25:20.

Measured voltage = 100(25/(20+25) = 56 V (2 sig fig)

(b) Another way to calculate combined resistance of two parallel resistors is to use the "1/x" key on the calculator with the formula: 1/R_t = 1/R_1 + 1/R_2.

Enter 30, press "1/x", press "+", enter 350, press "1/x", press "=", press "1/x": The answer is 27.63...

This divides with 20.

Measured voltage = 100(27.63/(20+27.63) = 58 V (2 sig fig).

(C) 1/Rt = 1/30 + 1/50000.

Rt = 29.982

Measured voltage = 100(29.982/(20+29.982) = 60 V (2 sig fig).

Answer 2

What is the EMF? The voltage given by the battery is it shown on the diagram??