$A$ cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when:

An electric lamp is marked $60\, W$, $230\, V$. The cost of $1\, kWh$ of energy is Rs. $1.25$. The cost of using this lamp for $8$ hours a day for $30$ days is Rs. ...............

$A$ piece of fuse wire melts when a current of $15\, A$ flows through it. With this current,if it dissipates $22.5\, W$,the resistance of the fuse wire will be .............. $\Omega$.

$A$ potential $V_0$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_1$. The wire is then cut into two equal halves and a potential of $V_0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$. The ratio $P_2: P_1$ is $\sqrt{x}: 1$. The value of $x$ is $.............$.

$A$ $220\, V$,$1000\, W$ bulb is connected across a $110\, V$ mains supply. The power consumed will be ............ $W$.

The resistance of a device component decreases as the current through it increases and it is described by the relation,$R = \frac{0.2 I}{I-4}$,where $I$ is the current. Determine the minimum power delivered. (Assume $I > 4$) (in $W$)