Answer:
b. 5.6 mA
Explanation:
As the voltmeter is connected to the resistor in parallel. The new resistance of the systems is
[tex]\frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{10000} + \frac{1}{47} = 0.0214[/tex]
R = 1/0.0214 = 46.78 Ω
The voltage is
[tex]U = r_2 I_1 = 47 * 1.2 = 56.4 V[/tex]
So the new current now of the system is
[tex]I_2 = U/R = 56.4/46.78 = 1.2056 A[/tex]
So about 1.2056 - 1.2 = 0.0056 A or 5.6mA is drawn away.