In the circuit shown in the figure below, the current supplied by the battery is:
 
1. \(2~\text A\) 
2. \(1~\text A\) 
3. \(0.5~\text A\) 
4. \(0.4~\text A\) 

If a resistance coil is made by joining in parallel two resistances each of 20$\Omega$. An emf of 2V is applied across this coil for 100 seconds. The heat produced in the coil is

(1) 20 J

(2) 10 J

(3) 40 J

(4) 80 J

$\mathrm{Two}<mspace\; width="1em"></mspace>\mathrm{battries}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{emf}<mspace\; width="1em"></mspace>{\mathrm{E}}_1,<mspace\; width="1em"></mspace>{\mathrm{E}}_2<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\mathrm{internal}<mspace\; width="1em"></mspace>\mathrm{resistance}<mspace\; width="1em"></mspace>{\mathrm{r}}_1,<mspace\; width="1em"></mspace>{\mathrm{r}}_2<mspace\; width="1em"></mspace>\mathrm{are}<mspace\; width="1em"></mspace>\mathrm{connected}<mspace\; width="1em"></mspace>in<mspace\; width="1em"></mspace>parallel.$
$<mspace\; width="1em"></mspace>\mathrm{The}<mspace\; width="1em"></mspace>\mathrm{effective}<mspace\; width="1em"></mspace>\mathrm{emf}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{circuit}<mspace\; width="1em"></mspace>\mathrm{across}<mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\mathrm{B}<mspace\; width="1em"></mspace>\mathrm{is}$

1.  $\frac{{\mathrm{E}}_1{\mathrm{r}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2{\mathrm{r}}_2}{{\mathrm{r}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{r}}_2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

2.  $\frac{{\mathrm{E}}_1{\mathrm{r}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2{\mathrm{r}}_1}{{\mathrm{r}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{r}}_2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

3.  $\frac{{\mathrm{E}}_1{\mathrm{r}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2{\mathrm{r}}_1}{{\mathrm{r}}_1<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{\mathrm{r}}_2}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

4.  $\frac{{\mathrm{E}}_1{\mathrm{r}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2{\mathrm{r}}_2}{{\mathrm{r}}_1<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>{\mathrm{r}}_2}$

The equivalent resistance across AB is :

1.  2R

2.  R/2

3.  4R/3

4.  3R/2

The equivalent resistance between points \(A\) and \(B\) in the circuit shown in the figure is:

In the circuit shown in the figure, the effective resistance between A and B is:

1.  2$\mathrm{\Omega }$

2.  4$\mathrm{\Omega }$

3.  6$\mathrm{\Omega }$

4.  8$\mathrm{\Omega }$

For the circuit shown in the figure, the value of R must be

1.  3$\mathrm{\Omega }$

2.  4$\mathrm{\Omega }$

3.  5$\mathrm{\Omega }$

4.  6$\mathrm{\Omega }$

Current I as shown in the circuit will be

1.  10 A

2.  $\frac{20}{3}<mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace>$

3.  $\frac{2}{3}<mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

4.  $\frac{5}{3}<mspace\; width="1em"></mspace>\mathrm{A}$

The current through the 5$\mathrm{\Omega }$ resistor is

(1) 3.2A                     

(2) 2.8A

(3) 0.8A                     

(4) 0.2A

In the figure, a carbon resistor has bands of different colours on its body as mentioned in the figure. The value of the resistance is:

 
1. \(2.2\) k\(\Omega\)
2. \(3.3\) k\(\Omega\)
3. \(5.6\) k\(\Omega\)
4. \(9.1\) k\(\Omega\)