The ammeter $A$ reads $2\, A$ and the voltmeter $V$ reads $20\, V$. The value of resistance $R$ is (Assuming finite resistances of ammeter and voltmeter).

Two batteries of e.m.f $4 \text{ V}$ and $8 \text{ V}$ 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

The equivalent resistance and potential difference between $A$ and $B$ for the circuit shown are respectively:

In terms of resistance $R$ and time $T$,the dimensions of the ratio $\frac{\mu}{\varepsilon}$ of the permeability $\mu$ and permittivity $\varepsilon$ is:

The terminal voltage of the battery, whose emf is $10 \, V$ and internal resistance is $1 \, \Omega$, when connected through an external resistance of $4 \, \Omega$ as shown in the figure is: (in $V$)

In the circuit shown,the resistance of the voltmeter is $10,000 \, \Omega$ and that of the ammeter is $20 \, \Omega$. The ammeter reading is $0.10 \, A$ and the voltmeter reading is $12 \, V$. Then $R$ is equal to ............. $\Omega$.