ADDITIONAL EXERCISES

A low voltage supply from which one needs high currents must have very low internal resistance. Why?

A high tension (HT) supply of, say, $6 \mathrm{kV}$ must have a very large internal resistance. Why?

Given $n$ resistors each of resistance $R,$ how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

Given the resistances of $1\mathrm{\Omega }, 2\mathrm{\Omega }, 3\mathrm{\Omega },$ how will you combine them to get an equivalent resistance of (i) $\left(11/3\right) \mathrm{\Omega }$ (ii) $\left(11/5\right) \mathrm{\Omega },$ (iii) $6 \mathrm{\Omega },$ (iv) $(6/11) \mathrm{\Omega }?$

Determine the equivalent resistance of networks shown below.

Determine the current drawn from a $12 \mathrm{V}$ supply with internal resistance $0.5 \mathrm{\Omega }$ by the infinite network shown below. Each resistor has $1 \mathrm{\Omega }$ resistance

The figure shows a potentiometer with a cell of $2.0 \mathrm{V}$ and internal resistance $0.40 \mathrm{\Omega }$ maintaining a potential drop across the resistor wire $\mathrm{A}\mathrm{B}.$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate currents up to a few $\mathrm{m}\mathrm{A}$) gives a balance point at $67.3 \mathrm{c}\mathrm{m}$ length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of $600 \mathrm{k}\mathrm{\Omega }$ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf $\mathrm{\epsilon }$, and the balance point found similarly, turns out to be at $82.3 \mathrm{c}\mathrm{m}$ length of the wire.

(a) What is the value $\mathrm{\epsilon }$?
(b) What purpose does the high resistance of $600 \mathrm{k\Omega }$ have? 
(c) Is the balance point affected by this high resistance?
(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of $1.0 \mathrm{V}$ instead of $2.0 \mathrm{V}$?
(e) Would the circuit work well for determining an extremely small emf, say of the order of a few $\mathrm{m}\mathrm{V}$ (such as the typical emf of a thermocouple)? If not, how will you modify the circuit?

The figure shows a $2.0 \mathrm{V}$ potentiometer used for the determination of internal resistance of a $1.5 \mathrm{V}$ cell. The balance point of the cell in open circuit is $76.3 \mathrm{c}\mathrm{m}$. When a resistor of $9.5 \mathrm{\Omega }$ is used in the external circuit of the cell, the balance point shifts to $64.8 \mathrm{c}\mathrm{m}$ length of the potentiometer wire. Determine the internal resistance of the cell.