Answer 1

case 1

Let the total resistance of the combination shown in the diagram below be R.
Here 500 Ω and 1000 Ω resistors are connected in series combination. Their net resistance is given by R = R₁+ R₂
Here:
R₁=500 Ω
R₂=1000 Ω
So:
R = 500 + 1000 = 1500 Ω
Thus, the total resistance of the combination is 1500 Ω. 

case 2

In the circuit diagram shown above, two 2 Ω resistors are connected in parallel combination.
Therefore: 

`1/R_-p=1/R_1+1/R_2` 

`1/R_p=1/2+1/2` 

`Ω1/R_p=2/2=1`Ω

 Thus the total resistance of the combination is 1 Ω. 

case 3

In the circuit diagram shown above, two 4 Ω resistors are connected in parallel combination. 

Therefore their resulant resistance given by 

`1/R=1/R1+1/R_2` 

`Here R_1=4`Ω 

`R_2=4`Ω 

So 

`1/R=1/4+1/4` 

`or 1/R=2/4` 

or R=2Ω 

In the above circuit diagram, this 2 Ω and 3 Ω resistors are connected in series combination. So, the total resistance of the circuit can be calculated as: 

`R=R_1+R_2` 

Here 

`R_1=2`â

`R_2=3`â       

So: 

R=2â+3â=5â 

Thus, the total resistance of the combination is 5 â„