Let m cells be connected in series and n such groups are connected in parallel
If the emf of each cell is E and internal resistance r then the total emf of m in series is mE and  the total internal resistance is mr. When n such groups are in parallel the effective internal resistance is mr/n. Then the current through an external resistance R is 
$l=\frac{mE}{R+\frac{mr}{n}}=\frac{mnE}{nR+mr}=\frac{mnE}{(\sqrt{nR}-{\sqrt{mr)}}^2+2\sqrt{mnRr}}$
Now i will be maximum if the denominator is the minimum i.e. if  nR=mr
Using  R= 3 ohm and r=1 ohm we have 3n = m
But  mn = 48
therefore $\frac{m\times m}{3}=48$
which gives m= 12
thus n=4

The given circuit is Wheatstone's bridge.The current through the galvanometer will be  zero if the bridge is balanced 
$\frac{P}{Q}+\frac{R}{S}$ where P = 2+3 =5 ohm, Q = 10 ohm and R=4 ohm
The value of S is given by 
$\frac{5}{10}=\frac{4}{S}$
or S= 8  ohm
thus the effective resistance of the parallel combination of 12 ohm and x ohm must be 8 ohm
therefore 
$\frac{1}{12}+\frac{1}{x}=\frac{1}{8}$
which gives x = 24 ohm

Resistance $R,(R+R)=\text{Â}2R,\text{Â}(R+R)=2R$ and R are in parallel 
Hence effective resistance is given by
$\frac{1}{R_{eff}}=\frac{1}{R}+\frac{1}{2R}+\frac{1}{2R}+\frac{1}{R}$
which gives $R_{eff}=\frac{R}{3}$

The emfs of cells connected in reverse polarity cancel each other. Hence cells marked 2, 3 and 4 together cancel the effect of cells marked 5, 6 and 7 and the circuit reduces to that shown in figure. Now cells 1 and 8 are in reverse polarity. Hence the voltmeter reading = 5 – 5 = 0 V

$Q=\frac{V^2}{R}$
But $R=\frac{pl}{\pi r^2}$
Therefore $Q=\left(\frac{\pi V^2}{p}\right)\frac{r^2}{1}$
Q is doubled if both l and r are doubled

copper is a conductor whereas Germanium is a semi conductor

The drift speed depends on A the cross sectional area of the conductor but the current is independent of A

The equivalent resistance between points A and B to the right of AB is 4 ohm. Therefore total resistance  =3+4+2= 9 ohm. Current I $=\frac{9}{9}$=1 A. This current is equally divided in the 8 ohm resistor between A and B and the remainder 8 ohm resistor. Hence current in  AC =0.5 A. This current is equally divided between the 8 ohm resistor in CD and the circuit to the right of CD. Therefore current in the 4 ohm resistor  = 0.25  A

In the steady state no current flows in the branch containing the capacitor .Thus the current say l flows in the  branches containing R and 2R. Applying Kirchoff's second rule to the loop abcdefa
2V-I(2R)-IR-V=0
$l=\frac{V}{3R}$
Potential drop across capacitor  =2V-V-l(2R)=$V−\frac{v}{3R}×\text{Â}2R$
$=V−\frac{2v}{3}=\frac{v}{3}$
 

The branches ABPQ and PQCD are a balanced Wheatstone's bridge. Therefore resistances between E and E and between F and G do not contribute and the circuit simplies to the one show in figure. The effective resistance $R_e$ between P and Q is given by
$\frac{1}{R_e}=\frac{1}{4R}+\frac{1}{2r}+\frac{1}{4R}$
which gives $R_e=\frac{2Rr}{(R+r)}$