The question involves analyzing the given circuit to find the current flowing through the ideal voltage source when:

1. The voltmeter is disconnected.
2. The voltmeter is connected.

Extracted Information:

1. The resistor $R = 50 \, \Omega$.
2. The voltmeter's internal resistance $R_v = 50 \, \text{M}\Omega$ (50 Megaohms).
3. The voltage of the source is not explicitly mentioned, so we'll use $E$ to denote it.

Case 1: Voltmeter disconnected

When the voltmeter is disconnected, the current is determined by the source voltage $E$ and the resistance $R$: $I_{\text{no voltmeter}} = \frac{E}{R}.$

Case 2: Voltmeter connected

When the voltmeter is connected, the voltmeter and $R$ form a parallel combination. The equivalent resistance of the circuit is: $R_{\text{eq}} = \frac{1}{\frac{1}{R} + \frac{1}{R_v}} = \frac{R \cdot R_v}{R + R_v}.$ Substituting the values $R = 50 \, \Omega$ and $R_v = 50 \, \text{M}\Omega$: $R_{\text{eq}} \approx 50 \, \Omega,$ since $R_v \gg R$, the resistance of the voltmeter hardly affects the equivalent resistance.

Thus, the current is approximately the same as in Case 1: $I_{\text{with voltmeter}} \approx \frac{E}{R}.$

Final Answer:

1. Current with the voltmeter disconnected: $I = \frac{E}{50}$.
2. Current with the voltmeter connected: $I \approx \frac{E}{50}$.

If more details or exact numerical values for $E$ are available, I can calculate specific results. Let me know!

Would you like additional clarifications?
Here are 5 related questions for deeper understanding:

1. What happens if the voltmeter resistance is comparable to $R$?
2. How would this change if $R$ were in series with another resistor?
3. Can you derive the percentage error introduced by the voltmeter?
4. How would the result change with a finite internal resistance in the voltage source?
5. What are the practical implications of such high-resistance voltmeters?

Tip: Always compare the voltmeter resistance with the circuit resistance to estimate its effect on the measurements!