Question

A circuit consists of a series combination of 6.50 −kΩ and 5.00 −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 5.00 −kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.(a) What potentialdifference does the voltmeter measure across the 5.00-kΩ resistor? (b) What is the true potential difference across thisresistor when the meter is not present? (c) By what percentage isthe voltmeter reading in error from the true potentialdifference?

Answer 1

Answer:

Part a)

$V = 16.95 Volts$

Part b)

$V = 21.74 V$

Part c)

$error = 22$%

Explanation:

Part a)

When voltmeter of 10 k ohm is used to measure the voltage across the given resistor then we will have

$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

$R_{eq} = \frac{5 \times 10}{5 + 10}$

$R_{eq} = \frac{50}{15} = \frac{10}{3} k$ ohm

now the current through the given cell is

$i = \frac{V}{r_1 + r_2}$

$i = \frac{50}{6.50 + \frac{10}{3}}\times 10^{-3}$

$i = 5.08 mA$

now voltage across 5 k ohm resistor is given as

$V = i R$

$V = (5.08 mA)(\frac{10}{3}\times 10^3)$

$V = 16.95 Volts$

Part b)

If there is not voltmeter connected across 5 k ohm

then current in the circuit is given as

$i = \frac{50}{6.5 + 5} \times 10^{-3} A$

$i = 4.35 mA$

Now voltage across 5 k ohm resistor

$V = iR$

$V = 4.35 \times 10^{-3} (5 \times 10^3)$

$V = 21.74 V$

Part c)

percentage error in the reading is given as

$error = \frac{V_1 - V_2}{V_1} \times 100$

$error = \frac{21.74 - 16.95}{21.74} \times 100$

$error = 22$%