Two batteries of e.m.f 4 V and 8 V with internal resistance $1 \Omega$ and $2 \Omega$ respectively are connected in a circuit with a resistance of $9 \Omega$ as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is $\mathrm{R}=9 \Omega$

With a resistance ' X ' connected in series with a galvanometer of resistance $100 \Omega$, it acts as a voltmeter of range $0-15 \mathrm{~V}$. To double the range, a resistance of $1500 \Omega$ is to be connected in series with ' X '. The value of ' X ' in ohm is

The current (I) drawn from the battery in the given circuit is

The value of the shunt resistance that allows $10 \%$ of the main current through the galvanometer of $99 \Omega$ is