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Debora [2.8K]
4 years ago
12

A circuit consists of a series combination of 6.00-kΩ and5.00-kΩ resistors connected across a 50.0-V battery havingnegligible in

ternal resistance. You want to measure the truepotential difference (that is, the potential difference without themeter present) across the 5.00-kΩ resistor using a voltmeterhaving an internal resistance of 10.00 kΩ. (a) What potentialdifference does the voltmeter measure across the 5.00-kΩresistor? (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?
Physics
1 answer:
Oksi-84 [34.3K]4 years ago
4 0

Answer:

(a) 17.87 V

(b) 22.73 V

(c) 21.38%

Solution:

As per the question:

Voltage of the battery, V = 50.0 V

Now, across the resistance  of 5.00k\Omega, a battery is connected in parallel with an internal resistance 10.00k\Omega

Thus the equivalent resistance of this parallel combination is given by:

\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10}

R_{eq} = \frac{5\times 10}{5 + 10} = 3.34\ k\Omega

the overall equivalent resistance across 50.0 V is:

R_{eq} = 3.34 + 6.00 = 9.34\ k\Omega

Now, the Current flowing in the circuit is given by:

I = \frac{V}{R_{eq}} = \frac{50.0}{9.34\times 10^{3}} = 5.35\times 10^{- 3}\ A = 5.35\ mA

(a) To calculate the potential difference across 5.00\ k\Omega, we use Ohm' law:

V = IR = 5.35\times 10^{- 3}\times 3.34\times 10^{3} = 17.87\ V

(b) To calculate the true potential difference in absence of the meter across the resistor:

When the meter is not connected, then both the resistors are in series and hence the resultant resistance is :

R' = 6.00 + 5.00 = 11.00 k\Omega

Now, the current in the circuit:

I' = \frac{V}{R'} = \frac{50}{11\times 10^{3}} = 4.55\times 10^{- 3}\ A = 4.55\ mA

Now, the true potential difference is:

V' = I'R = 4.55\times 10^{- 3}\times 5\times 10^{3} = 22.73\ V

(c) Error in the reading of the voltmeter is given by:

error = \frac{V' - V}{V'} = \frac{22.73 - 17.87}{22.73} = 0.2138

% error = 21.38 %

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