Cells, EMF, Kirchhoff's Rules & Wheatstone Bridge

A battery of emf $12\,\text{V}$ and internal resistance $3\,\Omega$ is connected to an external resistor. If the current in the circuit is $0.6\,\text{A}$, the voltage across the external resistor will be:

A uniform wire of resistance $12\,\Omega$ is cut into three pieces in the ratio of length $1 : 2 : 3$. Now the three pieces are connected to form a triangle. A cell of emf $8\,\text{V}$ and internal resistance $5\,\Omega$ is connected across the highest of the three resistors. The current through the circuit is:

A galvanometer of resistance $520\,\Omega$ is shunted with $20\,\Omega$ resistance to convert it into an ammeter. The resistance of the ammeter will be:

The current through a $\frac{4}{3}\,\Omega$ external resistance connected to a parallel combination of two cells of $2\,\text{V}$ and $1\,\text{V}$ emf and internal resistances of $1\,\Omega$ and $2\,\Omega$ respectively is ___________.

A cell of emf $1.1\,\text{V}$ and internal resistance $0.5\,\Omega$ is connected to a wire of resistance $0.5\,\Omega$. Another cell of the same emf is now connected in series with the intention of increasing the current but the current in the wire remains the same. The internal resistance of the second cell is ___________.

Kirchhoff's Second Law (Loop rule) is based on the law of conservation of:

In a potentiometer arrangement, a cell of $1.5\,\text{V}$ gives a balance point at $45.0\,\text{cm}$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $75.0\,\text{cm}$, what is the emf of the second cell?

A cell of constant emf is first connected to a resistance $R_1$ and then to $R_2$. If the power delivered in both cases is the same, then the internal resistance of the cell is:

To protect a galvanometer from possible damages due to large current,

which of the following should be connected to its coil?

A voltmeter of resistance $150\,\Omega$ is connected across A and B in the given circuit. The reading of voltmeter will be :

A potentiometer with a cell of 2.4 volt and internal resistance of $2\,\Omega$ maintains a potential drop across the resistance wire AB of length 2 meters and resistance $10\,\Omega$. A standard cell which maintains a constant emf of 'V' volt with internal resistance $0.2\,\Omega$ gives a balance point at $1.6\,\text{m}$ length of the wire. The value of emf of second (standard) cell (V) is :

To convert a galvanometer into an ammeter, one should connect :