Respuesta :
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 5.00[tex]k\Omega[/tex], a battery is connected in parallel with an internal resistance 10.00[tex]k\Omega[/tex]
Thus the equivalent resistance of this parallel combination is given by:
[tex]\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10}[/tex]
[tex]R_{eq} = \frac{5\times 10}{5 + 10} = 3.34\ k\Omega[/tex]
the overall equivalent resistance across 50.0 V is:
[tex]R_{eq} = 3.34 + 6.00 = 9.34\ k\Omega[/tex]
Now, the Current flowing in the circuit is given by:
[tex]I = \frac{V}{R_{eq}} = \frac{50.0}{9.34\times 10^{3}} = 5.35\times 10^{- 3}\ A = 5.35\ mA[/tex]
(a) To calculate the potential difference across [tex]5.00\ k\Omega[/tex], we use Ohm' law:
V = IR = [tex]5.35\times 10^{- 3}\times 3.34\times 10^{3} = 17.87\ V[/tex]
(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 [tex]k\Omega[/tex]
Now, the current in the circuit:
[tex]I' = \frac{V}{R'} = \frac{50}{11\times 10^{3}} = 4.55\times 10^{- 3}\ A = 4.55\ mA[/tex]
Now, the true potential difference is:
V' = I'R = [tex]4.55\times 10^{- 3}\times 5\times 10^{3} = 22.73\ V[/tex]
(c) Error in the reading of the voltmeter is given by:
[tex]error = \frac{V' - V}{V'} = \frac{22.73 - 17.87}{22.73} = 0.2138[/tex]
% error = 21.38 %