Calculating Internal Resistance Of A Non Ideal Voltmeter

Question

This question in should be simple, however I'm still struggling to wrap my head round it.

Question : "A non ideal voltmeter is connected across a DC battery of emf = 10V, and internal resistance. The voltmeter reads 9.8V. when the voltmeter is removed and a load of resistance of R = 9 ohms is instead connected across the battery, a current of 1A flows. What is the resistance of the voltmeter ?

Now I have drawn both circuits, and it is just a simple ohms law question. So calculating the current going across the voltmeter, then we can find it's resistance? But unsure how to use the second circuit to do so.

Answer 1

With the resistance connected, the battery is losing $1$ V with a current of $1$A. i.e. the battery has a series internal resistance of $1 \Omega$.
When the voltmeter is connected the battery looses $0.2$ V, which means that the voltmeter is absorbing $0.2$A, i.e. that it has an internal parallel resistance of $9.8/0.2= 49 \Omega$