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.
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 $9.8/0.2= 49 \Omega$