# A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligible internal r

Question

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 4.50 −kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

Part A: What potential difference does the voltmeter measure across the 4.50 kohm resistor?

PartB:What is the true potential difference across the resistor when the meter is not present?

Part C: By what percentage is the voltmeter reading in error from the true potential difference?

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1 year 2021-08-31T01:22:40+00:00 2 Answers 90 views 0

Part A with the effect of voltmeter = 16.15 V

Part B without the effect of voltmeter = 20.43 V

Part C Error = 20.94 %

Explanation:

Part A: What potential difference does the voltmeter measure across the 4.50kΩ  ohm resistor?

Since voltmeter has its own internal resistance, when it measures voltage across an element then that voltage is not true voltage across the element since the effect of voltmeter is also included. The internal resistance of the voltmeter becomes parallel to the resistance of element

Req = Rvm*R/Rvm+R

Where Rvm is the resistance of voltmeter and R is the resistance of the element R = 4.50 kΩ

Req = 4.50*10/4.50+10

Req = 3.103 kΩ

Now we calculate the current flowing in the circuit since it is series circuit, the current is same through all elements

I = V/Rtotal

Where Rtotal = 6.50 + 3.103 = 9.603 kΩ

I = 50/9.603

I = 5.206 mA

V = I*Req

V = 5.206*3.103

V = 16.15 V

Part B: What is the true potential difference across the resistor when the meter is not present?

Now we don’t want to include the effect of internal resistance of the voltmeter so

I = V/Rtotal

Now Rtotal = 6.50 + 4.50 = 11 kΩ

I = 50/11

I = 4.54 mA

V = I*R

V = 4.54*4.50

V = 20.43 V

Hence, as you can see the true voltage across the resistor is 20.43 V so internal resistance of the voltmeter shows reduced voltage across the resistor that was 16.15 V

Part C: By what percentage is the voltmeter reading in error from the true potential difference?

Percentage error = (20.43 – 16.15/16.15)*100 %

Percentage error = 20.94 %

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$$

$$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$$

Part C

The error in % is given by

$$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$$